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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
2s
Acceleration
= 8m/s
2s
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)
2s
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
t
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
2
3) Replace v in equation 2 by u +at
s = (u+at+u) t
2
s = (2u +at) t
2
s= ut + at2 ----------------------------- (3) ;
displacement
2
4
Replace u in equation 2 by v-at
s=
(v+ v-at) t
2
s=
(2v-at)t
2
s
= vt - at2
------------------------------ (4)
2
5. Make t the subject from equation (1)
above and substitute the t in equation (2).
From equation (1), t= v-u
a
Using equation 2; s= (v+u) t
2
s = (v+u)
(v-u) (difference of two squares)
2
a
s=v2-
u2
2a
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
2
s= (15+39)4
2
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
t
a= 13.98- 5.56
4
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
2
In the first second, where t=1,
15= u(1) + a(12)
2
15= u + a
2
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
2
40= u(2) + a(22)
2
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
2
1=u
(1) +a (1)2
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
2=
u(3)+ a(3)2
2
2=
3u + 9a
2
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
s
Solution
In
timet the body covers distance s=u(t) + a(t2)
2
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
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
4u
t=
s
2u
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
2
20= 20t- 5t2
2
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 5th second. Find the acceleration and the
distance covered in the 6th 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 5th 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.
|
a |
u |
v |
t |
s |
|
2 |
5 |
6 |
|
|
|
-4 |
17 |
15 |
|
|
|
|
4 |
|
2 |
20 |
|
-2 |
5 |
|
8 |
|
|
6 |
|
|
4 |
96 |
|
-2 |
17 |
|
|
-38 |
|
|
8 |
10 |
|
9 |
i)
t=0.5s, s=11/4m
ii)
t=0.5s , s=8m
iii)
a=6ms-2 , v=16ms-1
iv)
v=-11ms-1, s=-24m
v)
u=12ms-1, v=36ms-1
vi)
v=-21ms-1, t=19s
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