3 Constant Acceleration & Accelerated Motion
Resources
- Extra Help: A-Level Physics Tutor
- Extra Help: The Kinematic Equations
Equations Introduced and Used in this Topic: (All equations can be written and solved as both scalar and vector)
- [latex]a = \dfrac{\Delta v}{t}[/latex]
- [latex]\Delta v = v_f - v_i[/latex]
- [latex]v_f = v_i + at[/latex]
- [latex]d = v_it + \dfrac{1}{2} at^2[/latex]
- [latex]\vec{a} = \dfrac{\vec{\Delta v}}{t}[/latex]
- [latex]v = \dfrac{(v_i + v_f)}{2}[/latex]
- [latex]2ad = v_f^2 - v_i^2[/latex]
- [latex]d = \dfrac{(v_i + v_f) t}{2}[/latex]
Where…
- [latex]v[/latex] is average speed, commonly measured in metres/second (m/s) or kilometres/hour (km/h)
- [latex]\vec{v}[/latex] is average velocity, commonly measured in metres/second (m/s) or kilometres/hour (km/h) and includes a direction
- [latex]v_i[/latex] is initial speed, commonly measured in metres/second (m/s) or kilometres/hour (km/h)
- [latex]\vec{v_i}[/latex] is initial velocity, commonly measured in metres/second (m/s) or kilometres/hour (km/h) and includes a direction
- [latex]v_f[/latex] is final velocity, commonly measured in metres/second (m/s) or kilometres/hour (km/h) and includes a direction
- [latex]\Delta d[/latex] is distance travelled, commonly measured in metres (m), kilometres (km)
- [latex]\vec{\Delta d}[/latex] is change in displacement, commonly measured in metres (m), kilometres (km) and includes a direction
- [latex]a[/latex] is acceleration (deceleration is negative), measured in metres per second squared (m/s2)
- [latex]\vec{a}[/latex] is vector acceleration, measured in metres per second squared (m/s2) and includes a direction
- [latex]t[/latex] is time, commonly measured in seconds (s) or hours (h)
| Equations | a | vt | vi | d | t |
|---|---|---|---|---|---|
| [latex]v_t = v_i + at[/latex] | Not mentioned | ||||
| [latex]2ad = v_{r}^{2} - v_{i}^{2}[/latex] | Not mentioned | ||||
| [latex]d = v_{i}t + \dfrac{1}{2} at^2[/latex] | Not mentioned | ||||
| [latex]d = \dfrac{(v_i + v_t)t}{2}[/latex] | Not mentioned |
3.1 Average Acceleration
Extra Help: Acceleration
Acceleration (and deceleration) are two commonly experienced phenomenon by anyone who has rode in a vehicle, an elevator or more noticeably so on some ride at an amusement park. You notice acceleration as a change in either speed or velocity in either speeding up, slowing down or in changing the direction you are traveling. All of these changes to either your speed or velocity means that you are experiencing some sort of acceleration[1].
Quantifying acceleration and deceleration specifically refers to the rate[2] of the increase or the decrease in some objects speed (or velocity). This means that:
[latex]a = \dfrac{\Delta v}{t}[/latex] and since [latex]\Delta v = v_f - v_i[/latex], then [latex]a = \dfrac{v_f - v_i}{t}[/latex]
The acceleration equation is written the same for average vector acceleration except it has vector notion written over top of the vector variables.
There are a few distinctions in working with scalar and vector acceleration:
For Scalar Acceleration:
- Acceleration refers to the increase in some objects speed
- Deceleration refers to the decrease in an objects speed (Use a negative for this measure)
For Vector Acceleration:
- Acceleration refers to the change in velocity and can be a positive or a negative value
The units for acceleration are generally given in base SI metric form of m/s2 or as ms-2.
One can run into the form of units used in the UK that is given as m/s.
One of the rationalizations for using the UK units is due to the mental gymnastics of trying to understand time squared (s2).
Historically, the concept of acceleration is one that was influenced by observations of the experiments used by various natural philosophers. The oldest surviving records of the study of motion is that of Aristotle (384-322 BCE) who studied the motion of objects falling through water, his medium of choice that he needed to slow down the motion so it could be more easily studied. Water is around 784 times the density of air which significantly influenced all motions of any falling object. Any object falling through water reaches a maximum (terminal velocity) in a short order of time which means that it will then fall at a constant velocity downwards. Also, more massive objects will take a longer time to reach terminal velocity when falling downwards.
Aristotle interpreted these observations as follows:
- Heavy objects fall faster than lighter objects.
- The rate at which object fall depended on the density of the medium they were falling through.
- Objects move according to their ratios of their composition of Earth, Water, Air and Fire. All Natural motion therefore occurred to move all elements to their natural place in the universe.
While aware that objects could increase speed, Aristotelian scholars thought that speed was related to the distance separating these four elements and that as the distance decreased the speed increased. To Aristotelians time had no importance in these changes of speed. Aristotle did distinguish between motion that he considered natural and of motion caused by humans or other non-natural sources that he called violent (unnatural). This interpretation of agents acting on the motion of a body created the foundation of a concept called impetus[3], where the motion of a body that was thrown would continue in that motion until the impetus initially imparted to it wore off or was exhausted. This concept was refined nearly a millennium later by John Philoponus (490-570) in his modification of the concept of violent motion.
The Middle Ages witnessed the first of the many challenges that grew to eventually overthrow Aristotelian theory. The Persian polymath Avinenna (930-1037) is credited in moving away from pure Aristotelian concepts by claiming that the Philoponean theory of violent motion was motion that was dissipated by external agents such as friction from moving through its medium as the reason for the changes to violent motion of a body. In this, Avinenna predated Newton in arriving at his first law of motion. Persian science continued to develop concepts related to motion such as Ibn al-Haytham (965-1039) who linked the amount of acceleration related to gravitational acceleration. Abū Rayḥān Muḥammad ibn Aḥmad Al-Bīrūnī (973-1048) and Abu’l-Barakāt al-Baghdādī (1080–1165) both acted to construct the foundations of what later became know as Newton’s second law of motion that related the external force acting on an object resulted in the object accelerating. History abounds of the great advances coming from Islamic science which not for the total destruction of Baghdad and the House of Wisdom in 1258 during the Mongol invasion by Hulagu Khan could have predated the explosion of European Science by hundreds of years. Instead, from this loss, scientific advancement in Persia slowly diminished in its output and followed what had occurred previously in Egypt and Greece.
European science began to develop in the late Middle Ages from the rediscovery of the works of Aristotle and was well on its path to taking up the reigns left by Persian scientists but was stalled by the Black Death plague in 1348 that killed over one third of the population, reaching higher numbers in urban centres where the heart of science was emerging. It was later in the destruction of the through Byzantine empire by the Ottoman Turks, the infusion of Byzantine and Islamic scientists fleeing to safe refuge, mainly in Italy that the Renaissance started.
Galileo Galilei, perhaps one of the most famous of the Renaissance scientists is credited with accomplishments that he sometimes did not do. For instance, he was claimed to have climbed the Leaning Tower of Pisa and dropped two metal balls both light and heavy and to have proved that both fell at the same rate[4] and not differently as claimed by Aristotle. The actual experiment is known to have occurred in 1586 by Simon Stevin who dropped two lead spheres, (one that was ten times the weight of the other) from a church tower, 10 metres above ground.
Galileo did however, used an inclined ramp to slow down the acceleration of a falling object and was able to demonstrate that the distance fallen by an object without any friction influencing the fall was proportional to the square of the time the object fell. Specifically:
[latex]d = t^2[/latex]
The following examples all relate to the rate of change in an objects speed or velocity and the acceleration acting on the object. We will be using the following equations for these examples:
- [latex]a = \dfrac{\Delta v}{t}[/latex]
- [latex]\Delta v = v_f - v_i[/latex]
- Which can be expanded to the form:
[latex]a = \dfrac{v_f - v_i}{t}[/latex] - [latex]\vec{a} = \dfrac{\vec{\Delta v}}{t}[/latex]
- [latex]\vec{\Delta v} = \vec{v_f} - \vec{v_i}[/latex]
Which can be expanded to the form: - [latex]\vec{a} = \dfrac{\vec{v_f} - \vec{v_i}}{t}[/latex]
Example 3.1.1
If ICBC expects that a car can decelerate at 6.15 m/s2, what amount of time should be needed to brake to a stop if traveling at 90 km/h?
Solution
Data:
- [latex]a=-6.15\text{ m/}\text{s}^2[/latex]
- [latex]\Delta v = 90 \text{km/h}[/latex] to a full stop
First: Convert units and determine goal
- 90 km/h = 25 m/s
- [latex]t = \text{ find}[/latex]
Second: Find the change in speed
- [latex]\Delta v = v_f - v_i[/latex]
- [latex]\Delta v = 0 \text{ m/s} - 25 \text{ m/s} = -25 \text{ m/s}[/latex]
Third: Find the time
- [latex]a = \dfrac{\Delta v}{t}[/latex] or [latex]t = \dfrac{\Delta v}{a}[/latex]
- [latex]t = \dfrac{-25 \text{ m/s}}{-6.15 \text{ m/s}^2}[/latex] or 4.07 s (≈ 4.1 s)
Example 3.1.2
A common measure of acceleration (or deceleration) is to give values in terms of the average strength of the Earths gravity or g’s. If one g is given the value of 9.8 m/s2, what amount of time is needed for a jet accelerating at 3.0 g’s to reach a speed of 100 m/s from rest?
Solution
Data:
- [latex]a = 3.0 \text{ gs}\times 9.8 \text{ m/s}^2 \text{ or }29.4\text{ m/s}^2[/latex]
- [latex]\Delta v = 100 \text{ m/s} - 0 \text{ m/s or } 100 \text{ m/s}[/latex]
- [latex]t = \text{ Find}[/latex]
Solution:
- [latex]a = \dfrac{\Delta v}{t} \text{ or } t=\dfrac{\Delta v}{a}[/latex]
- [latex]t = \dfrac{100 \text{ m/s}}{29.4 \text{ m/s}^2}[/latex]
- [latex]t = 3.4 \text{ s}[/latex]
Example 3.1.3
What is the average acceleration of an electron that changes its speed from rest to a speed of 3.00 × 106 m/s in 6.00 µs?
Solution
Data:
- [latex]v = 3.00 \times 10^6 \text{ m/s}[/latex]
- [latex]t = 6.0 \mu s \text{ or } 6.0 \times 10^{-6} \text{ s}[/latex]
- [latex]a = \text{ Find}[/latex]
Solution:
- [latex]a = \dfrac{\Delta v}{t}[/latex]
- [latex]a = \dfrac{3.00 \times 10^6 \text{ m/s}}{6.00 \times 10^{-6} \text{ s}}[/latex]
- [latex]a = 5.00 \times 10^11 \text{ m/s}^2[/latex]
Example 3.1.4
If an startled driver has a response time of 1.80 s to step on the brakes of a car, what time is needed to stop in a school zone if traveling at 50 km/h and if the brakes decelerate the average car at 5.8 m/s2?
Solution
This solution has several parts.
First: Find time to respond
- The time to respond is 1.80 s or [latex]t_{\text{response}} = 1.80 \text{ s}[/latex]
Second: Find the time to brake
- First, [latex]50 \text{ km/h } \div 3.6 = 13.89 \text{ m/s}[/latex]
- Second, [latex]\Delta v = v_f - v_i[/latex] so [latex]\Delta v = 0 \text{ m/s } - 13.89 \text{ m/s } = -13.89 \text{ m/s}[/latex]
- Data: Find the time to brake
- [latex]a = - 5.8 \text{ m/s}^2[/latex]
- [latex]\Delta v = -13.89 \text{ m/s}[/latex]
- [latex]t_{\text{brake}} = \text{ Find}[/latex]
- Solution: Find the time to brake
- [latex]a = \dfrac{\Delta v}{t}[/latex] or [latex]t = \dfrac{\Delta v}{a}[/latex]
- [latex]t_{\text{brake}} = \dfrac{-13.89 \text{ m/s}}{-5.8 \text{ m/s}^2}[/latex]
- [latex]t_{\text{brake}} = 2.39 \text{ s}[/latex] (≈ 2.4 s)
Now, the total time is:
- [latex]t_{\text{total}} = t_{\text{response}} + t_{\text{brake}}[/latex]
- [latex]t_{\text{total}} = 1.8 \text{ s } + 2.4 \text{ s or } 4.2 \text{ s}[/latex]
Exercise 3.1
- A downhill skier accelerates from rest to a speed of 15 m/s in a time of 3.0 s. What would be this skiers average acceleration?
- A remote control car accelerates from 1.0 m/s to 6.8 m/s in 2.0 s. What was its average acceleration?
- Any object that is dropped from rest near the Earth’s surface will accelerate downwards at 9.8 m/s² (9.8 m/s² is known as 1 g or 1 gravity of acceleration). What should be the change in speed of an object that falls for 3.0 s? (Ignore air resistance.)
- A high end sports car accelerates at 5.0 m/s². What amount of time should it take to change its speed by 30 km/h?
- At high speeds, a particular automobile is capable of an acceleration of about 0.50 m/s². At this rate how much time does it take to accelerate from 90 to 110 km/h?
- What amount of time is required to bring a car traveling at 110 km/h to a full stop, if the brakes decelerate the car at 6.0 m/s²?
- A quarter mile dragster can go from rest to 432 km/h in 6.0 s. What is the average acceleration of this car? Convert this acceleration to gs (gravities).
- An electron changes its speed from rest to 2.76 × 107 m/s in a time of 1.25 ns (nanosecond). What is the average acceleration acting on the electron?
- If an average undistracted driver has a response time of 0.50 s to step on the brakes of a car, what time is needed to stop in a school zone when traveling at 30 km/h and if the brakes decelerate the average car at 6.0 m/s²?
Article to Read: WHO (2008) Speed Management Manual
3.2 Constant Accelerated Motion
Equations Introduced or Used for this Section:
- [latex]a = \dfrac{\Delta v}{t}[/latex]
- [latex]\Delta v = v_f - v_i[/latex]
- [latex]v_f = v_i + at[/latex]
- [latex]d = v_it + \dfrac{1}{2} at^2[/latex]
- [latex]\vec{a} = \dfrac{\vec{\Delta v}}{t}[/latex]
- [latex]v = \dfrac{(v_i + v_f)}{2}[/latex]
- [latex]2ad = v_f^2 - v_i^2[/latex]
- [latex]d = \dfrac{(v_i + v_f) t}{2}[/latex]
The equations you choose to solve an accelerated problem can be found by using the chart shown below. How we choose which equation to use depends for the most part on which variable is not given or asked for … We can call these variables “not mentioned”.
The chart below indicates which of the four constant acceleration equations you can use for the variable that is not mentioned.
| Equations | a | vt | vi | d | t |
|---|---|---|---|---|---|
| [latex]v_t = v_i + at[/latex] | Not mentioned | ||||
| [latex]2ad = v_{r}^{2} - v_{i}^{2}[/latex] | Not mentioned | ||||
| [latex]d = v_{i}t + \dfrac{1}{2} at^2[/latex] | Not mentioned | ||||
| [latex]d = \dfrac{(v_i + v_t)t}{2}[/latex] | Not mentioned |
For example, you are given the following data in a problem:
- [latex]a[/latex] = a (constant) value
- [latex]v_i[/latex] = a value
- [latex]v_f[/latex] = a value
- [latex]d[/latex] = asked to find
- [latex]t[/latex] = not mentioned
The equation to use is:
[latex]2ad = v_f^2 - v_i^2[/latex]
In a setting where there is a constant average acceleration, it is possible to combine several equations together to create four accelerated motion equations. Specifically:
- [latex]v_f=v_i+at[/latex]
- [latex]2ad = v_f^2 - v_i^2[/latex]
- [latex]d = v_it + \dfrac{1}{2} at^2[/latex]
- [latex]d = \dfrac{(v_i + v_f) t}{2}[/latex]
It is quite easy to show how two of these four equations are derived …
[latex]d = \dfrac{(v_i + v_f) t}{2}[/latex] is simply the expansion of [latex]d=vt[/latex] (originally given as [latex]v=\dfrac{d}{t}[/latex]) where the speed of the object has been replaced by taking the average of the initial and final speed (or velocity [latex]= \dfrac{(v_i + v_f) }{2}[/latex]
In this case you can see that:
For the equation [latex]d=vt[/latex] when you replace [latex]= \dfrac{(v_i + v_f) }{2}[/latex], you are left with [latex]d = \dfrac{(v_i + v_f) t}{2}[/latex]
[latex]v_f=v_i+at[/latex] is simply the rearrangement of the basic acceleration equation [latex]a=\Delta \dfrac{v}{t}[/latex] where [latex]\Delta v[/latex] is replaced with [latex]v_f-v_i[/latex] and [latex]a=\Delta \dfrac{v}{t}[/latex] is rewritten as [latex]at=\Delta v[/latex]
When combined you are left with [latex]at=v_f-v_i[/latex] more commonly written as [latex]v_f=v_i+at[/latex]
The other two equations [latex]d = v_it + \dfrac{1}{2} at^2[/latex] and [latex]2ad = v_f^2 - v_i^2[/latex] require a little more work to derive.
NOTE: For these four equations to work correctly, the acceleration must be uniform or constant.
Example 3.2.1
What is the average acceleration of a pickup truck that changes its speed from 60 km/h to 90 km/h in 6.0 s?
Solution
Data:
- [latex]a[/latex] = Find
- [latex]v_i[/latex] = 60 km/h or 16.7 m/s
- [latex]v_f[/latex] = 90 km/h or 25 m/s
- [latex]d[/latex] = not mentioned
- [latex]t[/latex] = 6.0s
Solution:
- Use [latex]v_f = v_i + a t \text{ or } a = \dfrac{v_f - v_i}{t}[/latex]
- [latex]a = \dfrac{25 \text{ m/s } - 16.7 \text{ m/s}}{6.0 \text{ s}}[/latex]
- [latex]a = 1.4 \text{ m/s}^2[/latex]
Example 3.2.2
If an object dropped from rest accelerates downwards at 9.8 m/s2 (one gravity or g), what distance should it have fallen after it fell for 4.0 s?
Solution
Data:
- [latex]a=9.8 \text{ m/s}^2[/latex]
- [latex]v_i=0\text{ m/s. Dropped}[/latex]
- [latex]v_f=\text{ not mentioned}[/latex]
- [latex]d=\text{ find}[/latex]
- [latex]t=4.0 \text{ s}[/latex]
Solution:
- Use [latex]d=v_it+\dfrac{1}{2}at^2[/latex]
- [latex]d = (0 \text{ m/s})(4.0 \text{ s}) + 0.5(9.8 \text{ m/s}^2)(4.0 \text{s})^2[/latex]
- [latex]d = 0 \text{ m} + 78.4 \text{ m}[/latex]
- [latex]d = 78.4 \text{ m}[/latex] (≈ 78 m)
Example 3.2.3
What distance does a vehicle travel when passing another vehicle if it constantly accelerates from 80 km/h to 115 km/h over 7.5 s?
Solution
Data:
- [latex]a[/latex] = not mentioned
- [latex]v_i=80 \text{ km/h or }22.2\text{ m/s}[/latex]
- [latex]v_f=115 \text{ km/h or }31.9\text{ m/s}[/latex]
- [latex]d=\text{ find}[/latex]
- [latex]t=7.5\text{ s}[/latex]
Solution:
- Use [latex]d = \dfrac{(v_i + v_f) t}{2}[/latex]
- [latex]d = \dfrac{(22.2 \text{ m/s } + 31.9 \text{ m/s})}{2} (7.5 \text{ s})[/latex]
- [latex]d =\dfrac{(54.1\text{ m/s})}{2} (7.5\text{ s})[/latex]
- [latex]d = 203 \text{ m}[/latex] (≈ 200 m)
Example 3.2.4
A texting driver takes 6.0 s to notice he (or she) needs to brake to a stop when traveling at 120 km/h. If we use ICBC’s estimation that the average vehicle can decelerate at 6.15 m/s2, what distance would this vehicle travel before coming to a complete stop?
Solution
Data for distracted driving:
- [latex]v=120\text{ km/h or }33.3\text{ m/s}[/latex]
- [latex]d[/latex] = find
- [latex]t=6.0 \text{ s}[/latex]
Solution for distracted driving:
- Use [latex]v = \dfrac{d}{t}\text{ or }d=vt[/latex]
- [latex]d = (33.3 \text{ m/s})(6.0 \text{s})[/latex]
- [latex]d = 200 \text{ m}[/latex]
Data for braking:
- [latex]a= -6.15 \text{ m/s}^2[/latex]
- [latex]v_i =120 \text{ km/h or } 33.3 \text{ m/s}[/latex]
- [latex]v_f=0 \text{ m/s. Stopped}[/latex]
- [latex]d[/latex] = find
- [latex]t[/latex] = not mentioned
Solution for braking:
- Use [latex]2ad = v_f^2 - v_i^2[/latex]
- [latex]2 (-6.15 \text{ m/s}^2) d = (0 \text{ m/s})^2 - (33.3 \text{ m/s})^2[/latex]
- [latex](-12.3 \text{ m/s}^2) d = -1111 \text{ m}^2/ \text{s}^2[/latex]
- [latex]d = \dfrac{-1111 \text{ m}^2 / \text{s}^2}{-12.3 \text{ m/s}^2}[/latex]
- [latex]d = 90.3 \text{ m}[/latex] (≈ 90 m)
Total distance driven:
- [latex]d_{\text{total}} = d_{\text{distracted}}+d_{\text{braking}}[/latex]
- [latex]d_{\text{total}} = 200 \text{ m} + 90.3 \text{ m}[/latex]
- [latex]d_{\text{total}} = 290 \text{ m}[/latex]
The above examples all relate to the constant acceleration of a body in a straight line. In all cases the speed of the object will be changing. It is also possible to have an object experience constant acceleration but not have any change in speed. This occurs for an object traveling in a circular path, such as a planet orbiting the Sun or a satellite orbiting a planet.
All motion depends on the direction of the acceleration that is acting on the body. If the acceleration is in the same direction (or opposite to the direction) that the body moves then the speed will be changing. If this acceleration is acting at 90° to the direction of motion, then the object will move in a circular path, remaining at the exact same speed but constantly changing its velocity. Motion in a circular path will be studied much later in this text.
Exercise 3.2
- A baseball pitcher requires 0.1 s to accelerate a ball from rest to a speed of 30 m/s. Calculate the average acceleration acting on this ball.
- An object starting from rest accelerates at 10 m/s2.
- What distance would it travel in 0.50 s?
- What speed does it reach in 0.50 s?
- A bicyclist has a speed of 6.0 m/s at the top of a hill, when he/she begins to accelerate at 2.0 m/s2.
- What amount of time is needed for this person to reach a speed of 15 m/s?
- What distance does this person ride downhill during this time?
- What distance does a sports car accelerating at 3.0 m/s2 need to go from:
- 0 m/s to 15 m/s?
- 15 m/s to 30 m/s?
- A small plane, touches down at 30 m/s and decelerates to a full stop 6.0 s later. How far did it travel in this time?
- A small car accelerates from rest at a rate of 5.2 m/s2 over a distance of 80 m before running into the rear of a pickup. At what speed was the small car traveling when the collision occurred?
- A small pickup required 6.0 s to come to a full stop from an initial speed of 108 km/h.
- What was the average deceleration acting on this pickup while braking?
- What amount of time should be needed to stop this pickup when traveling at 144 km/h? (Use the acceleration found in part (i))
- If it takes just 2.0 s to brake a car from 30 m/s to 20 m/s, what amount of time should be needed to brake the same car from 15 m/s to a full stop?
- A car starting at 36 km/h accelerates at 2.0 m/s2 for ½ a kilometre. What is the final speed reached by this car?
- What deceleration acts on a skier that stops in 40 m if initially traveling at 72 km/h?
- A truck accelerates from 40 km/h to 80 km/h in 5.0 s. What is its acceleration in m/s2 and what distance did it travel in this time?
- What is the expected distance a car will stop if traveling at 108 km/h and can brake with an average deceleration of 6.15 m/s2?
- A tourist is traveling at 110 km/h when she begins to read a sign on the side of the road. For what amount of distance was she unaware of what was on the road, if it took her 5 s to read this sign?
- You are driving 100 km/h and see a bunny run out in front of you. It takes you 0.75 s to apply your brakes. What distance do you travel in this time?
Exercise 3.3
- The average well maintained car should be able to decelerate at 0.6 g’s. What distance is required to bring this car to a full stop upon braking, when originally traveling at 120 km/h?
- A person who is properly constrained by a shoulder harness has a good chance of surviving a car collision if the deceleration does not exceed 30 g’s. Assuming uniform deceleration at this rate calculate the distance over which the front end of the car must be designed to collapse if a crash occurs at 100 km/h.
- Granny is 5.0 m away from safety. A killer tomato is 3.0 m behind Granny. Hearing a noise Granny turns and sees the killer tomato and begins to accelerate her wheelchair at 0.75 m/s2 towards safety. The killer tomato chases after her accelerating at 1.0 m/s2. Does Granny make it?
- A 100 m long train accelerates uniformly from rest. If the front of the train passes a railway worker 150 m down the track at a speed of 25 m/s, what will be the speed of the last car as it passes the worker?
Exercise 3.4.1: ICBC Road Sense Campaign
Calculate the deceleration in g’s for the car crash described in a ICBC promotion. Assume units are in kilometres per hour.
“120 to 0 in 3.1 seconds.”
Exercise 3.4.2: Star Trek Discovery Episode “The Vulcan Hello”
“Investigating a damaged satellite near a binary star system on the edge of Federation space, the crew members of the USS Shenzhou discover an object obscured from their sensors. After First Officer Michael Burnham volunteers to investigate the object, she finds an ancient, carved vessel. She is attacked by a Klingon, and when trying to escape, she accidentally kills him. A group of Klingons mourn the death of their soldier, dubbed the “Torchbearer”, before the outcast Voq volunteers to take his place. The Klingons, led by T’Kuvma, reveal themselves in a cloakable ship. T’Kuvma preaches to his followers of the Federations attempts to usurp the individuality of the Klingons and their culture, and plans to fulfill an ancient prophecy by uniting the 24 great Klingon houses as was once done by Kahless. Voq activates a beacon that summons the Klingon leaders. Burnham, desperate to prevent a war, attempts to fire on the Klingons first, against the wishes of Captain Philippa Georgiou. Burnham is arrested for mutiny.”
At the start of this episode Michael Burnham dons a space suit with a transport pack that takes her 2000 km in 6 minutes. The fastest way to travel this distance is to accelerate towards the spaceship for 3 minutes the first 1000 km) and then decelerate for the remaining 3 minutes to travel the remaining 1000 km to stop at the ship.
Question: Estimate the number of g’s acceleration Burhman experiences.
Exercise Answers
3.1 Average Acceleration
- 5.0 m/s2
- 2.9 m/s2
- 29 m/s
- 1.7 s
- 11 s
- 5.1 s
- 2 g’s
- 2.21 × 1016 m/s2
- 1.9 s
3.2 Accelerated Motion
- 300 m/s2
- 1.25 m (≈ 1.2 m)
- 5.0 m/s
- 4.5 s
- 47 m 4
- 37.5 m (≈ 37 m)
- 113 m (≈ 110 m)
- 90 m
- 29 m/s
- 5.0 m/s2, deceleration = 8.0 s to stop
- 3.0 s
- 46 m/s
- 5.0 m/s2
- 2.2 m/s2
- 83 m
- 73 m
- 153 m or 150 m
- 20.8 m or 21 m
3.3 Accelerated Motion
- 94 m
- 1.3 m
- Granny makes it!!! (by 0.35 s)
- 32 m/s
3.4.1 ICBC Road Sense Campaign
- Cars deceleration = 1.1 g’s
3.4.2 Star Trek Discovery Episode “The Vulcan Hello”
- 62 m/s2 or 6.3 g’s
Media Attributions
- “Focus Accident” by Johnathan Nightingale is licensed under CC BY-SA 2.0 licence.
- It is possible to experience acceleration without being able to detect it. ↵
- Rate is a measure of the change of something over the time it takes. ↵
- Impetus and the study of impulse & momentum are covered later in this text. ↵
- While claiming that Galileo did do the drop of the the two spheres, NASA astronaut David Scott in the Apollo 15 lunar mission dropped hammer and a feather on the Moon, crediting Galileo Galilei: The Apollo 15 Hammer-Feather Drop ↵