Question

A car is moving at 45 km/hr. A constant force acts on the car for 10 sec so that its velocity
becomes 63 km/hr. The distance travelled by car during this interval of 10 sec is

Answer 1

the laws of motion are:

   v = u + a t        --- (1)
   s = u t + 1/2 a t²    --- (2)
   v² = u² + 2 a  s    --- (3)

   s = (v + u) t / 2        --- (4)

since a constant force acts on the body, the acceleration of the body is constant.
  v = 63 kmph = 63 * 5/18 m/sec = 35/2  m/sec
  u = 45 kmph = 45 * 5/18 = 25/2 m/s

we apply the equation of motion (4),
   s = (35/2 + 25/2) * 10 / 2  m
       = 150  meters

   we could ALSO find acceleration by equation (1) and then find the distance / displacement using equation (2) or (3).

Answer 2

Here given,
u = 45 km/hr ; v = 63 km/hr and t = 10 s
u = 45 *5/18 m/sec = 12.5 m/sec
v = 63 * 5/18 m/sec = 17.5 m/sec
Now by first equation of motion, 
we will take out acceleration
v = u + at
=> at = v - u
=> a = (v - u)/t
=> a = (17.5 - 12.5)/10 = 5 / 10 m/sec²  = 0.5 m/sec²
now we have got a = 0.5 m/sec²
by third equation of motion,
v² - u² = 2as
=> s = (v² - u²) / 2a
=> s = {(17.5)² - (12.5)² } / 2*0.5
=> s = {306.25 - 156.25} / 1
=> s = 150 m