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### Constant acceleration

We are all familiar with the fact that a oto speeds up when we put our foot down on the accelerator. The rate of change of the velocity of a particle with respect lớn time is called its acceleration. If the velocity of the particle changes at a constant rate, then this rate is called the constant acceleration.

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Since we are using metres & seconds as our basic units, we will measure acceleration in metres per second per second. This will be abbreviated as m/s(^2). It is also commonly abbreviated as ms(^-2).

For example, if the velocity of a particle moving in a straight line changes uniformly (at a constant rate of change) from 2 m/s to 5 m/s over one second, then its constant acceleration is 3 m/s(^2).

Example

Let (t) be the time in seconds from the beginning of the motion of a particle. If the particle has a velocity of 4 m/s initially (at (t=0)) & has a constant acceleration of 2 m/s(^2), find the velocity of the particle:

when (t=1) when (t=2) after (t) seconds.

Draw the velocity–time graph for the motion.

Solution When (t=1), the velocity of the particle is (4 + 2 = 6) m/s. When (t=2), the velocity of the particle is (4 + 2 imes 2 = 8) m/s. After (t) seconds, the velocity of the particle is (4 + 2t) m/s.

Decreasing velocity

If the velocity of a particle moving in a straight line changes uniformly (at a constant rate of change) from 5 m/s lớn 2 m/s over one second, its constant acceleration is (-3) m/s(^2).

If a particle has an initial velocity of 6 m/s và a constant acceleration of (-2) m/s(^2), then:

when (t=1), the velocity of the particle is 4 m/s when (t=2), the velocity of the particle is 2 m/s when (t=3), the velocity of the particle is 0 m/s when (t=4), the velocity of the particle is (-2) m/s when (t=10), the velocity of the particle is (-14) m/s.

In general, the velocity of the particle is (6-2t) m/s after (t) seconds. The velocity–time graph for this motion is shown below; it is the graph of (v(t)=6-2t).

Over the first three seconds, the particle"s speed is decreasing (the particle is slowing down). At three seconds, the particle is momentarily at rest. After three seconds, the velocity is still decreasing, but the speed is increasing (the particle is going faster và faster).

Summary

If we assume that the rate of change of velocity (acceleration) is a constant, then the constant acceleration is given by

< extAcceleration = dfrac extChange in velocity extChange in time. >

More precisely, the constant acceleration (a) is given by the formula

where (v(t_i)) is the velocity at time (t_i). Since velocity is a vector, so is acceleration.

Example

A particle is moving in a straight line with constant acceleration of 1.5 m/s(^2). Initially its velocity is 4.5 m/s. Find the velocity of the particle:

after 1 second after 3 seconds after (t) seconds. Solution After 1 second, the velocity is (4.5 + 1.5 = 6) m/s. After 3 seconds, the velocity is (4.5 + 3 imes 1.5 = 9) m/s. After (t) seconds, the velocity is (4.5 + 1.5t) m/s.
Example

A car is travelling at 100 km/h (= dfrac2509) m/s, và applies its brakes lớn stop. The acceleration is (-10) m/s(^2). How long does it take for the oto to stop?

Solution

After one second, the car"s velocity is (dfrac2509 - 10) m/s. After (t) seconds, its velocity is

< v(t) = dfrac2509 - 10t ext m/s. >

The car stops when (v(t) = 0). Solving this equation gives

eginalign* dfrac2509 - 10t &= 0 \ t &= dfrac259. endalign*

The oto takes approximately 2.8 seconds to stop.

(In exercise 6, we will find out how far the car travels during this time.)

The constant-acceleration formulas for motion in a straight line

Throughout this section, we have been considering motion in a straight line with constant acceleration. This situation is very common; for example, a body toàn thân moving under the influence of gravity travels with a constant acceleration.

There are five frequently used formulas for motion in a straight line with constant acceleration. The formulas are given in terms of the initial velocity (u), the final velocity (v), the displacement (position) (x), the acceleration (a) & the time elapsed (t). Of course, they require consistent systems of units khổng lồ be used.

It is assumed that the motion begins when (t = 0), và that the initial position is taken as the origin, that is, (x(0) = 0).

The five equations of motion (v = u + at) (x = dfrac(u+v)t2) (x = ut + dfrac12at^2) (v^2 = u^2 + 2ax) (x = vt - dfrac12at^2)

Note. Each of the five equations involve four of the five variables (u, v, x, a, t). If the values of three of the variables are known, then the remaining values can be found by using two of the equations.

Deriving the constant-acceleration formulasThe first equation of motion

Since the acceleration is constant, we have (a = dfracv-ut). This gives the first equation of motion, (v = u + at).

The second equation of motion

The second equation,

says that displacement is obtained by multiplying the average of the initial và final velocities by the time elapsed during the motion. More simply:

< extDisplacement = extAverage velocity imes extTime taken. >

We can derive this equation using the fact that the displacement is equal khổng lồ the signed area under the velocity–time graph.

For the velocity–time graph on the left, the region under the graph is a trapezium. The displacement (x) is equal lớn the area of the trapezium, which is (dfrac12(u+v)t). So the second equation of motion holds in this case.

For the graph on the right, the displacement can be found by considering the two triangles between the graph & the (t)-axis. One of the triangles has positive signed area & the other has negative signed area.

Finding the displacement of a particle from the velocity–time graph using integration will be discussed in a later section of this module.

The third equation of motion

Substituting for (v) from the first equation into the second equation gives

eginalign*x &= dfrac(u+v)t2 \ &= dfrac(u+u+at)t2 \ &= dfrac2ut+at^22 \ &= ut + dfrac12at^2, endalign*

which is the third equation. Thus (x) is a quadratic in (t), and hence the graph of (x) against (t) is a parabola.

The fourth equation of motion

From the first equation, we have (t = dfracv-ua). Substituting this into the second equation gives

eginalign*x &= dfrac(u+v)t2 \ &= dfrac(u+v)(v-u)2a \ &= dfracv^2-u^22a. endalign*

Rearranging khổng lồ make (v^2) the subject produces the fourth equation: (v^2 = u^2 + 2ax).

The fifth equation of motion

From the first equation, we have (u = v-at). Using the second equation, we obtain

eginalign*x &= dfrac(u+v)t2 \ &= dfrac(v-at+v)t2 \ &= dfrac2vt-at^22 \ &= vt-dfrac12at^2, endalign*

which is the fifth equation.

Exercise 2

Derive the third and fifth equations of motion from a velocity–time graph. (For simplicity, you may assume that both (u) và (v) are positive.)

Vertical motion

Motion due to lớn gravity is a good context in which khổng lồ demonstrate the use of the constant-acceleration formulas. As discussed earlier, our two directions in vertical motion are up và down, and a decision has khổng lồ be made as to lớn which of the two directions is positive. Acceleration due khổng lồ gravity is a constant, with magnitude denoted by (g). In the following example, we take the upwards direction to lớn be positive và take (g = 10) m/s(^2).

Example

A stone is launched vertically upwards from ground màn chơi with the initial velocity 30 m/s. Assume that the acceleration is (-10) m/s(^2).

Find the time taken for the stone khổng lồ return to the ground again. Find the maximum height reached by the stone. Find the velocity with which it hits the ground. Sketch the position–time graph and the velocity–time graph for the motion. Find the distance covered by the stone from the launch to lớn when it returns to earth. Solution

In this example, we have (u=30) và (a=-10), with up as the positive direction.

Using the third equation of motion, (x = ut+dfrac12at^2), we have (x = 30t-5t^2). To find the time taken lớn return lớn the ground, we substitute (x=0) to obtain eginalign* 0 &= 30t-5t^2 \ 0 &= 5t(6-t). endalign* Therefore (t=0) or (t=6). Thus the time taken to lớn return lớn the ground is 6 seconds. We can find the maximum height using the fourth equation of motion, (v^2 = u^2+2ax). When the stone reaches the highest point, (v=0). So we have eginalign* 0 &= 900 - 20x \ x &= 45. endalign* Hence, the maximum height reached by the stone is 45 metres. The stone hits the ground when (t=6). Using the first equation of motion, < v = 30 - 10 imes 6 = -30 ext m/s. > From the third equation of motion, we have (x(t) = 30t-5t^2).

The maximum height reached by the stone is 45 m, & so the stone travels a distance of 90 m. Notes. Another way to lớn find the maximum height is by using the position–time graph. The graph of (x(t)) is a parabola with (t)-intercepts 0 & 6. Hence, the maximum value of (x(t)) occurs when (t=3). It follows that the maximum height reached by the stone is (x(3) = 30 imes 3 - 5 imes 3^2 = 45) metres. Looking at the velocity–time graph, we can see that the triangle above the (t)-axis has a signed area of 45 and the triangle below the (t)-axis has a signed area of (-45). Thus, when the stone has returned to ground level, it has travelled 90 m but its displacement is zero.

In the previous example, we took (-10) m/s(^2) as the approximate acceleration due khổng lồ gravity. A more accurate value is (-9.8) m/s(^2). Use this value in the following exercise.

Exercise 3

A man dives from a springboard where his centre of gravity is initially 12 metres above the water, and his initial velocity is 4.9 m/s upwards. Regard the diver as a particle at his centre of gravity, & assume that the diver"s motion is vertical.

Find the diver"s velocity after (t) seconds (up to lớn when he hits the water). Find the diver"s height above the water after (t) seconds (up to when he hits the water). Find the maximum height of the diver above the water. Find the time taken for the diver lớn reach the water. Sketch the velocity–time graph for this motion (up to when he hits the water). Sketch the position–time graph for this motion (up to lớn when he hits the water).
Exercise 4

A body toàn thân projected upwards from the đứng top of a tower reaches the ground in (t_1) seconds. If projected downwards with the same speed, it reaches the ground in (t_2) seconds. Prove that, if simply let drop, it would reach the ground in (sqrtt_1t_2) seconds.

Exercise 5

Example

A oto accelerates uniformly (constant acceleration) from 0 m/s khổng lồ 30 m/s in 12 seconds, & continues to accelerate at the same rate. Find

the acceleration the time it will take for the car to increase its velocity from 30 m/s to 80 m/s the distance the oto travels in the first 30 seconds of motion. Solution Use the first equation of motion, (v = u + at). Substituting (u=0), (v=30) and (t = 12) gives (30 = 12a). Hence, the acceleration is (a = dfrac52) m/s(^2). Use the first equation of motion again, with (u=30) và (v=80): < 80 = 30 + dfrac52t. > Thus (t = 20) seconds. Use the third equation of motion, (x = ut + dfrac12at^2). Substitute (u=0), (a=dfrac52) & (t=30): < x = dfrac12 imes dfrac52 imes 30^2 = 1125 ext metres. > The oto travels 1125 metres in the first 30 seconds of motion.
Exercise 6

A oto is travelling at 100 km/h (= dfrac2509) m/s. It brakes and slows down with an acceleration of (-10) m/s(^2). How far, lớn the nearest metre, does it go before it stops?

Exercise 7

A car accelerates from 0 km/h khổng lồ 100 km/h in 10 seconds, & continues for 40 seconds at 100 km/h. The driver then brakes strongly to stop in 38 metres.

Convert 100 km/h to lớn m/s. Find the constant acceleration of the oto for the first 10 seconds in m/s(^2). Find the total distance travelled by the car in metres. Find the acceleration for the braking phase in m/s(^2). How long does it take the oto to stop from when the brakes are first applied? Sketch a velocity–time graph for the motion of the car.