Question

A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Answer 1

Answer:

Acceleration = Change in Velocity/Time

Change in Velocity = 36-18 = 18 km/h=5 m/s

Time= 5 Seconds

Acceleration = 5/5= 1 m/s2

Equation of motion,s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5 s

a=1 m/s2

s= (5*5)+(1/2*1*5*5)

s=25+12.5    i.e., s=37.5 m

Explanation:

Answer 2

Answer:

Explanation:

Given,

initial velocity, u= 18km/h = 5 m/s

final velocity, v= 36km/h = 10 m/s

time, t= 5s

acceleration, a= ?

distance, s=?

now,

1. acceleration, a= (v-u)/t

a= (10-5)/5

a= 1 m/s

2. by 3rd equation of motion,

2as= v^2 - u^2

s= (v^2-u^2)/2a

s= (10^2-5^2)/2*1

s= 75/2

s= 37.5 m

Hope it helps...thank you