In similar manner, if we take the area above (or below) a velocity–time graph, we can obtain the displacement that the moving object undergoes.
The area above (or below) a velocity–time graph is the displacement that the moving object undergoes.
Let us now take an example involving an object which, like the car just considered, is moving in a straight line but, unlike the car, is moving with a changing velocity, the velocity increasing by the same amount every second.
A motion in which the speed or the velocity of an object changes by a constant amount every second is said to be moving with uniform
acceleration and the motion is referred to as a uniformly accelerated motion.
If the velocity is increasing with time, the object is said to be accelerating; if the velocity is decreasing with time, the object is said to be decelerating.
speed–time graph ❯
velocity–time graph ❯
Velocity is a vector. Speed is a scalar quantity.
uniform acceleration ❯ (a) 40 30 20 10 0 Time/s 0 1 2 3 4 V elocity v /m s –1 Figure 8.3
(a) Time/s t 0 V elocity v u (b) Time/s t 0 V elocity v u Figure 8.6
The general graph of velocity against time for a uniformly accelerated motion is shown in Figure 8.6 (a). It is a straight line and will always be straight with a
Worked example 8.2
Suppose the ball of Worked example 8.1 was released from a helicopter that was descending vertically at 8 m s–1.
(i) Draw a velocity–time graph for the motion of the ball.
(ii) What would be the ball’s velocity 4 seconds after being released?
Solution (i) 48 38 28 18 8 0 (b) Time/s 0 1 2 3 4 V elocity v /m s –1 Figure 8.4
(ii) As the ball has been released from a moving object, its initial velocity will be that of the moving object. Every second thereafter the velocity will increase by 10 m s–1 and so after 4 seconds the velocity would be
the sum of the initial velocity and the additional velocity due to the acceleration.
After 4 seconds the velocity is given by
v = initial velocity + increase due to acceleration
We write v = u + at (the first equation of motion), where v = the
velocity after time, t, commonly called ‘the final velocity’, u = the initial velocity, a = the constant acceleration and t = the time the motion lasted.
Substituting in the equation for this case, we have the velocity of the ball after 4 seconds,
v = 8 m s–1 + 10 m s–2 × 4 s = (8 + 40) m s–1 = 48 m s–1 MATHEMATICS: algebra – y = mx + c or y = c + mx ITQ1
The value of the acceleration due to gravity at the Earth’s surface is 10 m s–2.
(i) How long would it take an initial downward speed to increase from 10 m s–1 to 40 m s–1?
(ii) What is the acceleration for the motion represented by figure 8.5?
Time/s
Speed
Worked example 8.3
A stone, A, is released from a point X, high above the ground, Two seconds later another stone, B, is released from the same point (figure 8.7 (a)). (i) Draw a velocity–time graph for the motion of each stone.
(ii) Use the graphs to determine how far apart the stones would be 6 seconds after the first stone is released. Take the acceleration of free fall, g = 10 m s–2. (a) B X A 2 s 60 40 0 (b) Time, t/s Q Q′ O′ P A B 0 1 2 3 4 5 6 V elocity , v /m s –1 Figure 8.7 Solution
Figure 8.7 (a) shows the position of each stone at the moment of its release. (i) Figure 8.7 (b) shows the velocity–time graphs for the two motions.
Note that the slopes of the two graphs are the same. This must be so, since both stones are subject to the pull of gravity and will therefore have the same increase in velocity per second, or the same acceleration, 10 m s–2.
(ii) The areas under the respective graphs give the displacements of the two stones – see figure 8.7 (b).
The velocity of A after 6 seconds is
vA = u + at
= 0 + 10 m s–2 × 6 s
= 60 m s–1
So the displacement of stone A = the area under the velocity–time graph for stone A or the displacement of stone A from the starting
The phrase ‘acceleration due to gravity’ is sometimes referred to as ‘the acceleration of free fall’.
Graphs of motions with the same acceleration will have the same slope.
positive slope as long as the acceleration is positive or the velocity is increasing regularly with time (constant acceleration). If the motion was with constantly decreasing velocity, the acceleration would be negative, and the graph would again be a straight line but with a downward slope (see figure 8.6 (b)).
From the first equation of motion, we find that a = (v – u)/t, which is what we would expect if acceleration is defined as the change of velocity in unit time. It is important that you stick to the formula
acceleration = time taken for changechange of velocity = final moment (t) – initial moment (0)final velocity (v) – initial velocity (u)
If this formula produces a negative result, it means that the motion is a decelerated one, and that the object is slowing down with time.
MATHEMATICS: algebra – slope of a straight line graph
= (y 2 – y 1)/t
ITQ2
Sketch a speed–time graph to show each of the following motions: (i) A ball moving very fast and with
constant speed that is suddenly stopped.
(ii) A humming-bird that flies horizontally at constant speed, stops for a while and then flies vertically upwards at twice the speed.
(iii) An ant that moves quickly forwards, stops for a while, then crawls very slowly backwards. (iv) A ball dropped from a certain
height, rebounding twice on the ground at half the speed of impact each time.
We will now go back to the case we considered at the start of the chapter concerning cars A and B.
Suppose that at the very moment car A is at the point X travelling to the right, car B passes the same point X in the opposite direction travelling at the same speed of 20 m s–1. As time continues, car B gets closer and closer to the
origin, O. Table 8.2 shows how the distance of car B from O (the displacement of B) varies with time measured from the start of the observation.
The distance–time graphs (figure 8.8 (a)) for the two cars travelling between O and X will have exactly the same shape, but not the displacement–time graphs (figure 8.8 (b)). The distance graphs must look exactly the same, since the two cars both have the same speed and so will have covered the same distance in the same time. You will have noticed that they both have the same slope – this is no surprise, since we are only concerned with the magnitude or the rates of change of distance, and not the direction.
+ (a) Time, t/s A B 0 0 1 2 3 4 5 6 7 8 9 10 Distance travelled + (b) Time, t/s A B 0 0 1 2 3 4 5 6 7 8 9 10 Displacement + (c) Time, t/s A B 0 0 1 2 3 4 5 6 7 8 9 10 Speed + (d) Time, t/s A B 0 – 1 2 3 4 5 6 7 8 9 10 V elocity
sA = the area under graph A = (12 base × height) of ΔOPQ = 12 × 6 s × 60 m s–1
= 180 m
The velocity of stone B after falling for 4 s (2 seconds later than A) = 4 s × 10 m s–2 = 40 m s–1
So the displacement of B from the starting point, X, after 4 s, is (12 × base × height) of ΔO/PQ/ = 1
2 × 4 s × 40 m s–2
= 80 m
Since the displacements both have the same direction (downwards from the point of release), the separation of the stones will therefore be given by
sA – sB = 180 m – 80 m
= 100 m
The stones will therefore be separated by 100 m.
MATHEMATICS: geometry – area of a triangle
= 12 (base) × (height)
ITQ3
A car starts from rest and accelerates at a constant rate. After travelling a distance of 200 m, the speed of the car is 80 km h–1. What was the acceleration
of the car? Table 8.2
Time interval from the start for car B, t/s
Distance of car B from the origin, O (displacement of car B), s/m 5 100 6 80 7 60 8 40 9 20 10 0
Note the difference in the slopes of the two displacement graphs, however. Graph A rises uniformly, showing an increasing displacement as time passes. Graph B falls uniformly, since the displacement (the separation of the car from O) gets less and less with time at a uniform rate. Notice, also, that since the slope of graph A rises, showing a positive rate of change with time, or velocity, the associated velocity is positive. Graph B shows a negative slope, indicating a negative velocity. We expect the velocity of car B to be negative since it is moving to the left starting point, X. This is a good example of the dependency of velocity on displacement and also on direction.
Thus, the slope of graph A, defined as
final displacement – initial displacement time taken for change
gives (100 – 0)m(5 – 0)s = +20 m s–1. However, the slope of graph B = (0 – 100) m
(10 – 5) s
= –20 m s–1.
Note that the displacements and the times at which they occur must correspond. Thus the coordinates for A are (0 s, 0 m) initially and (5 s,100 m) finally, and for B (5 s, 100 m) initially and (10 s, 0 m) finally.
You will notice that, although the magnitude of the velocity is the same, 20 m s–1, as we would expect, the sign is negative, indicating that, since velocity
is a vector, it is in the opposite direction to the velocity of car A. So the velocity of car B is negative because its motion is to the left. This does not depend on starting point. Displacement does and so the displacement of both cars is always positive (and so above the time axis), as both cars are at all times to the right of the origin, O.
Let us now take a look at the speed–time and the velocity–time graphs for the cars. For car A, the velocity is to the right and so, as is the custom, we will take this velocity as positive. For car A the velocity is positive (since the slope is positive), whereas car B’s velocity is going to be negative, as shown by the negative slope of the velocity–time graph and by the motion of B to the left. Figure 8.8 (c) shows the speed–time graphs for the two cars. These are the same for both cars. When we consider the velocity–time graphs (see figure 8.8(d)) for A, however, the velocity is at a constant value of +20 m s–1 for the
first 5 s, and for B it is a constant value of –20 m s–1 for the second 5 s. Note that
when we multiply the constant positive velocity of +20 m s–1 by the time of
travel, 5 s, we are really multiplying a constant velocity by a time and getting a displacement of +100 m to the right from A’s starting point. When we do the same for B using its starting point, X, we obtain a negative displacement to the left of X. When we add together these two displacements, one relative to O and the other relative to X, we obtain +100 m + (–100 m) = 0, showing that the overall displacement of cars A and B is zero. Remember that both displacements were taken relative to the origin, O. Compare this addition of two vectors with that of the addition done in figure 5.1 (b).
This study of motion is called kinematics. What we have done applies to motion in a straight line only. It has not taken into account the factors or conditions that produce a change in velocity. Why does a body move as it does? Under what conditions will it speed up or slow down? Why does a body move in a circle? A study of these matters is a study of Newton’s laws. It is a study of dynamics to which we now turn.
MATHEMATICS: adding vectors
ITQ4
An object is projected vertically up into the air from a balcony high above the ground. It comes to rest momentarily after 3 seconds and then returns to a point 10 m below its point of projection at a time t 0 seconds later.
For the entire motion, draw: (a) a velocity–time graph; (b) a speed–time graph; and (c) an acceleration-time graph which shows that acceleration can be considered a vector quantity. Show the quantities given where possible on your graphs.
kinematics ❯
Figure 8.9 Sir Isaac Newton.