Question

A car accelerates uniformly from 18 km h to 36km h in 5 sec calculate acceleration and distance cover by car in that time​

Answer 1

Answer:

Initial speed = u = 18 km/hr = 5 m/s Final sped = v = 36 km/hr = 10 m/s acceleration a = (v-u)/t = (10-5)/5 = 1 m/s2 Distance travelled S = u t + (1/2) a t2 = 5×5 + (1/2)×1×52 = 37.5 m

Explanation:

Answer 2

Answer:

Acceleration = [(final velocity - initial velocity)/time]

1 km/hr = (5/18) m/s

Therefore, 18 km/hr = [18*(5/18)] m/s

= 5 m/s.

And, 36 km/hr = [36*(5/18)] m/s

= 10 m/s

Therefore, acceleration = [(10 - 5)/5 sec]

Thus, acceleration = 1 m/s2.

Equation of motion can be given as:

s=ut+(1/2)at²

u=18km/h=5m/s

t=5s

a=1m/s

s=(5×5)+( 2/1×1×5×5)

s=25+12.5

s=37.5m

Explanation:

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