Question

A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate
(i) acceleration in m/s2
(ii) distance covered by the bus in metres during this interval.

Answer 1

Answer:-

(i) acceleration = 0.5 m/s²

(ii) distance covered = 175 m

Given:-

v is the final velocitiy.

u is the initial velocity.

a is the acceleration.

t is the time.

S is the distance.

Initial speed u = 72km/h = 72 × 5 / 18 m/s = 20m/s

Final speed v = 54km/h = 54 × 5 / 18m/s  = 15m/s

Time = 10 seconds

(i) acceleration in m/s²

Acceleration a = (v - u)/t

⇒  20-15 / 10

⇒ 5 /10

⇒  0.5 m/s²

Acceleration a = 0.5 m/s²

(ii) distance covered by the bus in metres during this interval.

To find distance travelled, we use S = ut + ½ at²

Distance travelled  S = ut + ½ at²

⇒ 20² = 15² + 2 × 0.5 × S

⇒ 400 = 225 + 1×S

⇒ 400 - 225 = S

⇒ S = 175 m

Distance travelled  S = 175 m

Explanation:-

The three equations of motion

• v = u + at ;
• s = ut + (1/2) at²
• v²  = u²  + 2as

Answer 2

acceleration is taken out by law v=u + at

72 k/hr= 20m/s

54 k/hr=15m/s

20=15 + 10a

-10a= -5

a= -5/-10

a=0.5 m/s^2

Distance covered is to be found by 2as= v^2-u^2

2X0.5s= (20)^2 -( 15)^2

s=400 - 225

s=175 m