Question

A car accelerates uniformly from 20 km/h to 35km/h I 5 seconds .i] Calculate the acceleration.ii] The distance covered by that time

Answer 1

here initial velocity is 20 km/h
and final velocity is 35km/h

changing km/h into m/s

20×1 km by 1h

1km is 1000m and 1h is 3600 s

20×1000 divided by 3600s
we get
5.5m/s
converting final velocity
35 ×1000 divided by 3600s
we get
8.3 m/s
a= v- u by t
2.8 by 5
we get5.6m/s square

Answer 2

• What is Acceleration...???

• Change in the velocity of an object per unit time.

• Acceleration a = v-u/t

• S.I Unit : m/s²

• Here, u = initial velocity and v = final velocity.

         a = acceleration and t = time.

i) u = 20 km/h = 5.5 m/s

 v = 35 km/h = 9.7.. m/s

 t  =  5 second

 a = v-u/t

 a = 9.7 - 5.5 / 5

 a = 4.2 / 5

 a = 0.84 m/s²

ii) The Distance covered :

   s = ut + 1/2 at²

   Here, s = distance travelled by the object.

         t = time

         a = uniform acceleration.

         s = 5.5 × 5 + 1/2 ( 0.84 ) ( 5 )²

         s = 27.5 + 1/2 × 0.84 × 25

         s = 27.5 + 10.5

         s =  38m.