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Physics Form 4 Chapter 2 Force and Motion I

Relationship between Displacement,
EXAMPLE 2.9
Velocity, Acceleration and Time
Hisham drove his car from home with a uniform
Motion with Uniform Velocity acceleration and achieved a velocity of 15.0 m s –1
1. For an object moving with uniform velocity, v, in 5.0 seconds. What is the
its displacement, s after time, t is given by:
(a) acceleration of Hisham’s car?
(b) displacement of Hisham’s car 5.0 seconds
after starting the journey?
where v is the uniform velocity of the (c) velocity of Hisham’s car at t = 4.0 s?
object.
(d) velocity of Hisham’s car when he has
Motion with Uniform Acceleration travelled 20.0 m from the starting point?
1. The equations of motion of an object with Solution
uniform acceleration, a, are as follows: Summary of information:
Chapter
2 Displacement, s is given by: Initial velocity, u = 0 m s –1 –1
s = (Average velocity) × time
Final velocity, v = 15.0 m s
=  Initial velocity + final velocity  × time Time taken, t = 5.0 s
2
=  u + v t (a) Acceleration, a = v – u ............................. 
2 t
Hence, = 15.0 – 0
5.0
.................................  = 3.0 m s –2

1
2
2. Acceleration, a = v – u ............................... (b) Displacement, s = (u + v)t .....................
t
= 1 (0 + 15.0)(5.0)
Rearranging equation, 2
= 37.5 m
.......................................  (Can also be solved by equation )
3. Substitue equation  into equation . (c) Hisham’s velocity at t = 4.0.
1
s = (u + u + at)t v = u + at.....................................................
2 = 0 + 3.0 (4.0)
Simplifying .................. = 12.0 m s –1
(d) If the velocity of Hisham’s car after moving
4. Rearrange equation  into 20.0 m is v.
v – u With the information u = 0 m s –1
t = a a = 3.0 m s –2
and substitue into equation , s = 20.0 m
1
hence, s = (u+v) v – u  and using the equation 
2
a
v = u + 2as

2
2
2
2
2
= v – u 2 Therefore v = 0 + 2 (3.0 × 20.0)
2a = 120.0
Simpliying, Hence, v = 10.95 m s –1
................................... 

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