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1.2Uniformly accelerated motion: the equations of kinematics

Defining velocity in non-uniform motion

How is velocity defined when it is not constant? We have to refine what we did in Section 1.1. We now define the average velocity total displacement divided by total time as:

v¯=Δ s Δ t

where Δs is the total displacement for the motion and Δt the total time taken. We would like to have a concept of velocity at an instant of time, the (instantaneous) velocity the rate of change of position; it is a vector. We need to make the time interval Δt very, very small. The instantaneous velocity is defined as:

v=lim Δ t→0 Δ s Δ t

In other words, instantaneous velocity is the rate of change of position. This definition implies that velocity is the gradient of a position-against-time graph.

Consider Figure 1.6a. Choose a point on this curve. Draw a tangent to the curve at the point. The gradient of the tangent line is the meaning of v=lim Δ t→0 Δ s Δ t and, therefore, also of velocity.

Figure 1.6: a In uniformly accelerated motion the graph of position against time is a curve. b The gradient (slope) of the tangent at a particular point gives the velocity at that point.

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In Figure 1.6b the tangent is drawn at t = 3.0 s. We can use this to find the instantaneous velocity at t = 3.0 s.

The gradient of this tangent line is:

25−1.0 5.0−1.0=6.0 m s−1

To find the instantaneous velocity at some other instant of time we must take another tangent and we will find a different instantaneous velocity. At the point at t = 0 it is particularly easy to find the velocity: the tangent is horizontal and so the velocity is zero.

From now on we drop the word instantaneous and refer just to velocity. The magnitude of the velocity is known as the (instantaneous) speed the magnitude of the instantaneous velocity.

When the velocity changes we say that we have acceleration. The average acceleration is defined as a¯=Δ v Δ t. So for a body that accelerates from a velocity of 2.0 m s−1 to a velocity of 8.0 m s−1 in a time of 3.0 s the average acceleration is a¯=8.0−2.0 3.0=2.0 m s−2..

To define instantaneous acceleration or just simply acceleration, we use the same idea as for velocity. We let the time interval Δt get very small and define acceleration as:

Acceleration the rate of change of velocity; it is a vector is the rate of change of velocity. a=lim Δ t→0 Δ v Δ t. It is the gradient of a velocity-against-time graph. Acceleration is a vector.

Uniformly accelerated motion means that the acceleration is constant: the graph of velocity against time is a non-horizontal straight line (Figure 1.7). In equal intervals of time the velocity changes by the same amount.

Figure 1.7: In uniformly accelerated motion the graph of velocity against time is a straight line with non-zero slope.

Figure 1.8: This fighter jet is accelerating.

When the acceleration is positive, the velocity is increasing (Figure 1.8). Negative acceleration means that v is decreasing.

For constant acceleration there is no difference between instantaneous acceleration and average acceleration.

Suppose we choose a time interval from t = 0 to some arbitrary time t later. Let the velocity at t = 0 (the initial velocity) be u and the velocity at time t be v. Then:

a=Δ v Δ t=v−u t−0

which can be re-arranged to:

v=u+a t

For uniformly accelerated motion, this formula gives the velocity v of the moving object t seconds after time zero, given that the initial velocity is u and the acceleration is a.

Worked Example 1.2

A particle has an initial velocity 12 m s−1 and moves with a constant acceleration of −3.0 m s−2. Determine the time at which the particle stops instantaneously.

Answer

At some point it will stop instantaneously; that is, its velocity v will be zero.

We know that v = u + at.

Substituting values gives:

0=12+(−3.0)×t 3.0 t=12

Hence t = 4.0 s.

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Consider the graph of velocity against time in Figure 1.9. Imagine approximating the straight line with a staircase. The area under the staircase is the change in position since at each step the velocity is constant. If we make the steps of the staircase smaller and smaller, the area under the line and the area under the staircase will be indistinguishable and so we have the general result that:

KEY POINTS

The area under the curve in a velocity against time graph is the change in position; in other words, the displacement.

From Figure 1.9 this area is (the shape is a trapezoid):

Δ s=(u+v 2)t

Figure 1.9: The straight-line graph may be approximated by a staircase.

But v = u + at, so this becomes:

Δ s=(u+u+a t 2)t=u t+1 2 a t 2

So we have two formulas for position in the case of uniformly accelerated motion (recall that Δs = s − si):

s=s i+(u+v 2)t or Δ s=(u+v 2)t s=s i+u t+1 2 a t 2 or Δ s=u t+1 2 a t 2

We get a final formula if we combine s=s i+u t+1 2 a t 2 with v = u + at.

From the second equation, we find t=v−u a.

Substituting in the first equation we get:

s−s i=u v−u a+1 2 a(v−u a)2=u v a−u 2 a+1 2 v 2 a−u v a+1 2 u 2 a=v 2−u 2 2 a

This becomes:

v 2=u 2+2 a(s−s i) or v 2=u 2+2 a Δ s

Usually si = 0 so this last equation is usually written as v2 = u2 + 2as.

This formula is useful to solve problems in which no information about time is available.

In summary, for motion along a straight line with constant acceleration:

v=u+a t Δ s=(u+v 2)t Δ s=u t+1 2 a t 2 v 2=u 2+2 a Δ s

Graphs of position against time for uniformly accelerated motion are parabolas (Figure 1.10). If the parabola ‘holds water’ (concave up) the acceleration is positive. If its concave down, the acceleration is negative.

Figure 1.10: Graphs of position s against time t for uniformly accelerated motion. a Positive acceleration. b Negative acceleration.

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Exam tip

Table 1.2 summarises the meaning of the gradient and area for the different motion graphs.

G raph of … | G radient | A rea
position against time | velocity |
velocity against time | acceleration | change in position, i.e. displacement
acceleration against time |  | change in velocity

Table 1.2: Information that can be derived from motion graphs

The equations of kinematics v=u+a t, Δ s=u t+1 2 a t 2, Δ s=(u+v 2)t, 2=u 2+2 a Δ s
where u = initial velocity, v = final velocity, s = final position, a = acceleration, t = time and Δs = displacement can be used only for motion on a straight line with constant acceleration. (If the initial position is zero, Δs may be replaced by just s.)

Check yourself 2

The graph in Figure 1.11 shows the variation of position with time.

Figure 1.11

Indicate the intervals of times for which the acceleration is:

apositive

bnegative

czero.

Worked Example 1.3

A particle has an initial velocity 2.00 m s−1. Its acceleration is a = 4.00 m s−2. Find its displacement after 10.0 s.

Answer

Displacement is the change of position, i.e. ∆s = s − s i. We use the equation:

Δ s=u t+1 2 a t 2=2.00×10.0+1/2×4.00×10.0 2=220 m

Worked Example 1.4

A car has an initial velocity of u = 5.0 m s−1. After a displacement of 20 m, its velocity becomes 7.0 m s−1. Find the acceleration of the car.

Answer

Here, ∆s = s − s i = 20 m.

So, use v 2 = u 2 + 2a∆s to find a.

7.0 2=5.0 2+2 a×20 24=40 a

Therefore, a = 0.60 m s−2.

Worked Example 1.5

A body has an initial velocity of 4.0 m s−1. After 6.0 s the velocity is 12 m s−1. Determine the displacement of the body in the 6.0 s.

Answer

We know u, v and t. We can use:

Δ s=(v+u 2)t

to get:

Δ s=(12+4.0 2)×6.0 Δ s=48 m

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A slower method would be to use v = u + at to find the acceleration:

12=4.0+6.0 a⇒a=1.333 m s−2

Then use the value of a to find ∆s:

Δ s=u t+1 2 a t 2=4.0×6.0+1 2×1.333×36=48 m

Worked Example 1.6

A ball, X, starts moving to the right from rest with a constant acceleration of 2.0 m s−2. 2.0 s later, a second ball, Y, starts moving to the right with a constant velocity of 9.0 m s−1. Both balls start from the same position.

Determine the position of the balls when they meet. Describe what is going on.

Answer

Let the two balls meet t s after X starts moving.

The position of X is:

s X=1 2 a t 2=1 2×2.0×t 2=t 2

The position of Y is:

s Y=9(t−2)

(The factor t – 2 must be considered because, after t s, Y has actually been moving for only t − 2 seconds.)

These two positions are equal when the two balls meet, and so:

t 2=9(t−2)

Solving the quadratic equation we find t = 3.0 s and 6.0 s. At these times the positions are 9.0 m and 36 m.

At 3.0 s Y caught up with X and passed it. At 6.0 s X caught up with Y and passed it. (Making position–time graphs here is instructive.)

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Worked Example 1.7

A particle starts out from the origin with velocity 10 m s−1 and continues moving at this velocity for 5 s. The velocity is then abruptly reversed to −5 m s−1 and the object moves at this velocity for 10 s.

For this motion find:

athe change in position, i.e. the displacement

bthe total distance travelled

cthe average speed

dthe average velocity

Answer

The problem is best solved using the velocity–time graph, which is shown in Figure 1.12.

Figure 1.12

aThe initial position is zero. So, after 5.0 s, the position is 10 × 5.0 m = 50 m (the area under the first part of the graph). In the next 10 s, the displacement changes by −5.0 × 10 = −50 m (the area under the second part of the graph). So the change in position (the displacement) = 50 − 50 = 0 m.

bTake the initial velocity as moving to the right. The object moved towards the right, stopped and returned to its starting position. (We know this because the displacement was 0.) The distance travelled is 50 m moving to the right and 50 m coming back, giving a total distance travelled of 100 m.

cThe average speed is 100 15=6.7 m s−1 .

dThe average velocity is zero, since the displacement is zero.

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Worked Example 1.8

An object with an initial velocity 20 m s−1 and initial position of −75 m experiences a constant acceleration of −2 m s−2. Sketch the position–time graph for this motion for the first 20 s.

Answer

Use the equation s=s i+u t+1 2 a t 2. Substituting the values we know, the position is given by s = −75 + 20t − t 2. This is the function we must graph. The result is shown in Figure 1.13.

Figure 1.13

At 5 s the object reaches the origin and overshoots it. It returns to the origin 10 s later (t = 15 s). The furthest it gets from the origin is 25 m. The velocity at 5 s is 10 m s−1 and at 15 s it is −10 m s−1. At 10 s the velocity is zero.

A special acceleration

Assuming that we can neglect air resistance and other frictional forces, an object thrown into the air will experience the acceleration of free fall the acceleration, g, due to the pull of the earth on a body; g = 9.8 m s−2 near the surface of the earth while in the air. This is an acceleration caused by the attraction between the earth and the body. The magnitude of this acceleration is denoted by g. Near the surface of the earth g = 9.8 m s−2. The direction of this acceleration is always vertically downwards. (We will sometimes approximate g to 10 m s−2.)

Worked Example 1.9

An object is thrown vertically upwards with an initial velocity of 20 m s−1 from the edge of a cliff that is 25 m from the sea below, as shown in Figure 1.14.

Determine:

athe ball’s maximum height

bthe time taken for the ball to reach its maximum height

cthe time to hit the sea

dthe speed with which it hits the sea

(You may approximate g to 10 m s−2.)

Figure 1.14: A ball is thrown upwards from the edge of a cliff.

Answer

We have motion on a vertical line so we will use the symbol y for position (Figure 1.15a). We take the direction up to be positive. The zero for position is the ball’s initial position.

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Figure 1.15: Diagrams for solving the ball’s motion. a Displacement upwards is positive. b The highest point is the zero of position. Here we take displacement downwards to be positive.

aThe quickest way to get the answer to this part is to use v 2 = u 2 − 2gy. (The acceleration is a = −g.) At the highest point v = 0, and so:

0=20 2−2×10 y⇒y=20 m

bAt the highest point the object’s velocity is zero. Using v = 0 in v = u − gt gives:

0=20−10×t t=20 10=2.0 s

cThere are many ways to do this. One is to use the displacement arrow shown in blue in Figure 1.15a. Then, when the ball hits the sea, y = −25 m. Now use the formula y=u t−1 2 g t 2 to find an equation that only has the variable t:

−25=20×t−5×t 2 t 2−4 t−5=0

This is a quadratic equation. Using your calculator you can find the two roots as −1.0 s and 5.0 s. Choose the positive root to find the answer t = 5.0 s.

Another way of looking at this is shown in Figure 1.15b. Here we start at the highest point and take the direction down to be positive. Then, at the top y = 0, at the sea y = +45 m and a = +10 m s−2. Now, the initial velocity is zero because we take our initial point to be at the top.

Using y=u t+1 2 g t 2, we find:

45=5 t 2⇒t=3.0 s

This is the time to fall to the sea. It took 2.0 s to reach the highest point, so the total time from launch to hitting the sea is:

2.0+3.0=5.0 s.

dUse v = u − gt and t = 5.0 s to get v = 20 − 10 × 5.0 = −30 m s−1. The speed is then 30 m s−1.

(If you prefer the diagram in Figure 1.15b for working out part c and you want to continue this method for part d, then you would write v = u + gt with t = 3.0 s and u = 0 to get v = 10 × 3.0 = +30 m s−1.)

Check yourself 3

A ball is thrown vertically upwards with a speed of 20 m s−1 and another vertically downwards with the same speed and from the same height. What interval separates the landing times of the two balls? (Take g = 10 m s−2.)

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TEST YOUR UNDERSTANDING

5The initial velocity of a car moving on a straight road is 2.0 m s−1. It becomes 8.0 m s−1 after travelling for 2.0 s under constant acceleration. Find the acceleration.

6A car accelerates from rest to 28 m s−1 in 9.0 s. Find the distance it travels.

7A particle has an initial velocity of 12 m s−1 and is brought to rest over a distance of 45 m. Find the acceleration of the particle.

8A particle at the origin has an initial velocity of −6.0 m s−1 and moves with an acceleration of 2.0 m s−2. Determine when its position will become 16 m.

9A plane starting from rest takes 15.0 s to take off after speeding over a distance of 450 m on the runway with constant acceleration. Find the take-off velocity.

10A particle starts from rest with constant acceleration. After travelling a distance d the speed is v. What was the speed when the particle had travelled a distance d 2?

11A car is travelling at 40.0 m s−1. The driver sees an emergency ahead and 0.50 s later slams on the brakes. The deceleration of the car is 4.0 m s−2.

aFind the distance travelled before the car stops.

bCalculate the stopping distance if the driver could apply the brakes instantaneously without a reaction time.

cCalculate the difference in your answers to a and b.

dAssume now that the car was travelling at 30.0 m s−1 instead. Without performing any calculations, state whether the answer to c would now be less than, equal to or larger than before. Explain your answer.

12Two balls are dropped from rest from the same height. One of the balls is dropped 1.00 s after the other. Air resistance is ignored.

aFind the distance that separates the two balls 2.00 s after the second ball is dropped.

bExplain what happens to the distance separating the balls as time goes on.

13A particle moves in a straight line with an acceleration that varies with time as shown in Figure 1.16. Initially the velocity of the object is 2.00 m s−1.

Figure 1.16

aFind the maximum velocity reached in the first 6.00 s of this motion.

bDraw a graph of the velocity against time.

14The graph in Figure 1.17 shows the variation of velocity with time of an object. Find the acceleration at 2.0 s.

Figure 1.17

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15Your brand-new convertible is parked 15 m from its garage when it begins to rain. You do not have time to get the keys, so you begin to push the car towards the garage. The maximum acceleration you can give the car is 2.0 m s−2 by pushing and 3.0 m s−2 by pulling back on the car. Find the least time it takes to put the car in the garage. (Assume that the car and the garage are point objects.)

16A stone is thrown vertically up from the edge of a cliff 35.0 m from the sea (Figure 1.18). The initial velocity of the stone is 8.00 m s−1.

Figure 1.18

D etermine:

athe maximum height of the stone

bthe time when it hits the sea

cthe velocity just before hitting the sea

dthe distance the stone covers

ethe average speed and the average velocity for this motion

17A ball is thrown upwards from the edge of a cliff with velocity 20.0 m s−1. It reaches the bottom of the cliff 6.0 s later. Take g = 10 m s−2.

aDetermine the height of the cliff.

bCalculate the speed of the ball as it hits the ground.

18A toy rocket is launched vertically upwards with acceleration 3.0 m s−2. The fuel runs out when the rocket reaches a height of 85 m.

aCalculate the velocity of the rocket when the fuel runs out.

bDetermine the maximum height reached by the rocket.

cDraw a graph to show the variation of

ithe velocity of the rocket

iithe position of the rocket from launch until it reaches its maximum height

dCalculate the time the rocket takes to fall to the ground from its maximum height.