Question

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

Answer 1

Answer:

a=3.6

Explanation:

a=v-u/t

a=36-18/5

a=18/5

a=3.6m/s2

V2=u2+2as

(36)^2-(18)^2=2*3.6*S

(36+18)(36-18)=2*3.6*S

54*18/2*3.6=s

S=540/4

S=135m

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.