"Solving Problems on Motion of Bodies Moving in a Straight Line with Constant Acceleration"

ACCELERATION

Acceleration is a measure of the rate at which velocity changes. The SI units is ms^-2 or m/s². It is a vector quantity. Negative acceleration is called deceleration or retardation.

Acceleration = Final velocity-Initial velocity

Time

Example 1. A car moving with velocity of 4m/s along a straight road increases its velocity to 12m/s in 2seconds. Calculate the acceleration.

Solution

Initial velocity=4m/s

Final velocity=12m/s

Time= 2s

Acceleration= final velocity-initial velocity

Time

Acceleration= 12m/s-4m/s

Acceleration = 8m/s

Acceleration= 4m/s².

Example 2. A train is moving on a straight rail with constant acceleration of 15ms^-2 and has a velocity of 4ms^-1 at one point. Find its velocity 2 seconds later.

Solution

Acceleration= 15ms^-2

Initial velocity=4ms^-1

Final velocity=?

Time=2s

Acceleration=Final velocity-Initial velocity

Time

15ms^-2= Final velocity- 4ms^-1 ( multiply both sides by 2)

30ms^-1=Final velocity-4ms^-1

Final velocity=30ms^-1+ 4ms^-1

Final velocity=34ms^-1.

EXERCISE

1) A particle has velocity of 40m/s and in 4seconds, its velocity changes to 20m/s. Calculate its acceleration. What is the other name of this acceleration?

2) A particle moving in a straight line has constant acceleration. At one point, its velocity is -25ms^-1 and 5 seconds later its velocity is 10ms^-1. Find its acceleration.

3) A car moving with velocity of 36ms^-1 experiences an acceleration of -9ms^-1 in 4 seconds. Calculate its velocity at the end of the 4seconds.

4) A car moving in a straight road has acceleration -7ms^-2 with velocity of 12ms^-1. Find its velocity: a)8s later, b)12s later, c) 15s later

EQUATIONS FOR MOTION WITH CONSTANT ACCELERATION.

When the motion of a body is being considered, the letters u,v,a,t, and s usually have the following meanings:

u=initial velocity.

v=final velocity.

s= final displacement.

t=time taken or time interval.

a=acceleration.

Consider a car travelling in a straight line. If its initial velocity is 5m/s and 3 seconds later its velocity is 11m/s , the car is said to be accelerating at 2m/s².

Using the formula for calculating acceleration, we will derive equations that will ease calculations.

1) Acceleration (a)= final velocity(v)-initial velocity(u)

Time (t)

a= v-u

at=v-u

v=u+at -----------------(1) ; final velocity

2) Average Velocity= increase in displacement

Time taken

v + u = s

2 t

s= ( v +u)t ----------------------------(2) ; displacement

3) Replace v in equation 2 by u +at

s = (u+at+u) t

s = (2u +at) t

s= ut + at² ----------------------------- (3) ; displacement

4 Replace u in equation 2 by v-at

s= (v+ v-at) t

s= (2v-at)t

s = vt - at² ------------------------------ (4)

5. Make t the subject from equation (1) above and substitute the t in equation (2).

From equation (1), t= v-u

Using equation 2; s= (v+u) t

s = (v+u) (v-u) (difference of two squares)

2 a

s=v²- u²

2as= v² –u²

v² =u² +2as ------------------(5)

This 5 equations will greatly help us to answer questions involving straight line motion with constant acceleration. They will free us from visiting first principle each time a problem is given. Note that sometimes, the speed will be given in kmh^-1 to be converted to cms^-1 or ms^-1. Note; 1kmh^-1= 5 ms^-1 , 1ms^-1= 18kmh^-1

18 5

Example 3. A body starts at 15ms^-1 and accelerates at 6ms^-2 for 4 seconds. Calculate the final speed and distance covered.

Solution

u=15m/s

v=?

a=6m/s²

t=4s

a) v= u+at

v= 15m/s +(6m/s²)(4s)

v=39m/s is the final speed

b) s= (v+u)t

s= (15+39)4

s =108m

By using equations 4 and 5 above give the same results.

Example 4 . A car travelling at a velocity of 20kmh^-1accelerates for 4 seconds to reach a velocity of 50kmh^-1. Calculate the acceleration in ms^-1 and the distance covered in 4s.

Solution

u= 20km/h

v=50km/h

t=4s

using v= u +at

u= 20X 5 m/s = 100 m/s = 5.56m/s

18 18

v= 50 X 5 m/s = 250 m/s = 13.89m/s

18 18

From v= u + at

a= v –u

a= 13.98- 5.56

a=2.08m/s²

Example 5. A particle moving in a straight line with constant acceleration covers 15m and 25m in two successive seconds. Find the acceleration and velocity with which it enters the first 15m distance.

Solution

Let u be the velocity with which it enters the first 15m distance, ‘a’ the acceleration and s the distance covered.

Using s= ut + at²

In the first second, where t=1,

15= u(1) + a(1²)

15= u + a

30= 2u+ a

a + 2u= 30 --------------- (1)

In the complete period of 2 seconds, that is t=2s,

Total distance s= 15m + 25m= 40m.

From s= ut + at²

40= u(2) + a(2²)

80= 4u +4a

4a+ 4u = 80

a +u = 20 -------------- (2)

Solving equations (1) and (2) above simultaneously as follows:

Equation (1) - Equation (2) gives u=10m/s.

Substituting u in equation (2) gives a= 10m/s^-2

Example 6.A particle moving in a straight line with constant acceleration covers two successive distances of 1m in 1s and 2s.Find the acceleration and initial speed.

Solution

In the first 1m, the time taken is 1s. Using s=ut+1at²

1=u (1) +a (1)²

2=u + a

a + 2u = 2 ------------- (1)

In the complete journey,

Total distance= 1m+1m= 2m

Total time taken= 1s+2s=3s

Using s=ut+at²

2= u(3)+ a(3)²

2= 3u + 9a

4=6u+9a

9a+6u=4 ------------- (2)

Solving equations 1 and 2 simultaneously gives

a=- 1 ms^-2 and u= 7ms^-1

3 6

Example 7. A body moving with uniform acceleration begins with speed u in successive time t cover distances of s and 2s. Show that its acceleration is 4u²

Solution

In timet the body covers distance s=u(t) + a(t²)

2s= 2ut + at²-------- (1)

In the complete journey

Total time used = t +t=2t

Total distance covered= s +2s=3s

3s= u (2t) +a(2t)²

6s= 4ut +4at² -------------- (2) or 3s= 2ut +2at² ----------------- (2^’)

Equation (1) X 4 gives, 8s= 8ut +4at² ------------------- (3)

Equation (3) – Equation (2) gives 2s = 4ut

4ut = 2s

t= 2s

t= s

Equation 2^’- Equation1

s= at² -------------- (4)

s = a(s)²

(2u)²

s= as²

4u²

4u²s = as²

a= 4u²s

s²

a= 4u²

s as required.

Example 7. A particle moving in a straight line with uniform acceleration of -5m/s² has an initial velocity of 20ms^-1. Calculate time taken to get a displacement of: a) 20m, b)-20m from the starting point.

Solution

a) u=20m/s, a= -5m/s², s= 20m

Using s= ut + at²

20= 20t- 5t²

40=40t-5t²

5t²-40t+40=0

Solving using the quadratic formula gives t= (4- ...) s, t=(4+...)S

Exercises.

1.) A car starts from rest and accelerates uniformly for 8 seconds and reaches velocity of 40m/s. find the acceleration and distance covered.( 5m/s, 160m)

2.) A lorry travelling in a straight line with acceleration -15m/s^-2 has an initial velocity of 30m/s. When the velocity does becomes zero?( 2s)

3.) In two successive seconds, a uniformly accelerating body travels 3m and 7m respectively. Find its acceleration and initial velocity. (4m/s², 1ms^-1)

4.) A particle in a straight line with constant acceleration and staring from rest covers 36 m in the 5^th second. Find the acceleration and the distance covered in the 6^th second. (8m/s² , 44m)

5.) A train moving in a straight line with constant acceleration covers distances 3m and 9m respectively in the first 2 seconds of its motion. Find the acceleration and the distance travelled in the 5^th second of his motion. ( 6ms^-2, 27m)

6.) A car moving in a straight line with constant deceleration covers successive distance of 100m in 5s and 10s. Calculate the initial speed, deceleration and the further time needed for the car to come to rest. (70/3 ms^-1, 4/3ms^-2 , 2.5s)

7.) A cyclist covers a distance of 54m accelerating at 10ms^-2 for 6s. Find the initial and final velocities. (-21ms^-1, 39ms^-1)

8.) From a speed of 144km/h, a car decelerates uniformly to rest, covering a distance of 80m. Calculate the deceleration and the time taken. (10ms^-1, 4s)

9) Given that a, u, v, t, s have units ms^-2, ms^-1 , ms^-1, s, m respectively. Fill the blanks on the table below.