Question

A motorcycle,travelling east starts from rest ,moves in a straight line with a constant acceleration and covers a distance of 64m in 4s.calculate (a)its acceleration
(b)it's final velocity
(c)at what time the motorcycle had covered half the total distance.
(d)what distance the motorcycle had covered in half the total time

Answer 1

Answer:

(a) 8 m/s^2

(b) 32 m/s

(c) 2.82 s ( approx.)

(d) 16 m

Explanation:

Given:

constant acceleration ( => we can use the equations of motion)

u = 0 m/s

s = 64 m

t = 4s

Since he is travelling in one direction only, every vector quantity will be +ve

Average velocity = Total displacement / total time

Here, since the motorcycle has linear motion in one direction, distance = displacement = 64m

Avg. velocity = 64 /4

= 16 m/s

Also, for a given time, avg. velocity = ( u + v) /2

=> 16 = ( 0 + v) /2

=> v = 32 m/s

=> final velocity = 32 m/s

a = ( v - u) /t

= 32/4

= 8 m/s^2

=> acceleration = 8 m/s^2

Now ,

s = 1/2 x 64

= 32m

a = 8 m/s^2

u = 0 m/s

t = ?

s = ut + 1/2at^2

=> 32 = 0 + 1/2 x 8 x t^2

=> 32 = 4t^2

=> t^2 = 8

=> t = 2.82 seconds ( approx)

The motorcyclist covers half the total distance in 2.82 seconds.

Now,

t = 1/2 x 4

= 2 seconds

a = 8 m/s^2

u = 0 m/s

s = ?

s = ut + 1/2at^2

=> s = 1/2 x 8 x 2 x 2

=> s = 16m

The motorcyclist covered a distance of 16m in exactly 1/2 the total time.