Question

A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find :
(a) the speed acquired.
(b) the distance travelled.​

Answer 1

Given:-

• Initial velocity,u = 0 m/s [ Bus starts from rest ]

• Acceleration,a = 0.1 m/s²

• Time taken,t = 2 minutes

To be calculated:-

calculate:(a) the speed acquired and (b) the distance travelled.

Solution:-

Time taken = 2 minutes

= 2 × 60 seconds

= 120 seconds

Now,

( a )

From the first equation of motion

v = u + at

★ Substituting the values in the above formula,we get:

v = 0 + 0.1 × 120

v = 1/10 × 120

v = 1 × 12

v = 12 m/s

Thus,the speed acquired by bus is 12 m/s.

( b )

From the second equation of motion,

s = ut + 1/2 at²

★ Substituting the values in the above formula,we get:

s = 0 × 120 + 1/2 × 0.1 × ( 120 )²

s = 0 + 1/2 × 1/10 × 14400

s = 1/2 × 1 × 1440

s = 1/2 × 1440

s = 720 m

Thus,the distance covered by the bus is 720 m.

Answer 2

Given

u=0 a= 0.1m/s^2 t = 2 min = 120 s

from the equation of motion v = u + at

= 0 + 0.1 * 120

= 0+12

= 12

therefore v =12m/s

distance travelled -- S = ut + 1/2at^2

=0*120 + 1/2 * 0.1 * 120 * 120

= 0 + 720

= 720 m

total distance travelled = 720 m