Question

A car accelerates uniformly from 18km/hr to 36 km/hr in 5 sec.calculate the distance covered by the car in 5 sec.

Answer 1

 Acceleration = Change in Velocity/Time 

Change in Velocity = 36-18 = 18 km/h=5 m/s 
Time= 5 Seconds 
Acceleration = 5/5= 1 m/s^2 

Equation of motion,s=ut+(1/2)at^2 
u=18 km/h=5 m/s 
t=5 s 
a=1 m/s^2 
s= (5*5)+(1/2*1*5*5) 
s=25+12.5
 i.e.,
s=37.5 m 

Hope you are clear with my explanations

Answer 2

u= initial velocity= 18 km/h = 18 x 5/18= 5 m/s
v= final velocity = 36 km/h = 36 x 5/18 = 10 m/s
t= time = 5 seconds
a = acceleration = (v-u)/t = (10-5)/5 = 5/5 = 1 m/s²
s= distance = ?

using second equation of motion,

s= ut+ 1/2 at²

we get,

s= 5 x 5 + 1/2 x 1 x 5²
s = 25 + 12.5
s= 37.5 m

∴ the distance covered is 37.5m