Question

A body starts from the rest with constant
acceleration and travels 120 meter in 8th second.
The distance travelled by body in 4 s is​

Answer 1

Answer:

30 meters.

Step-by-step explanation:

Initial Velocity (u) = 0 (because the body started from rest)

Distance Travelled (s₁) = 120 m

Time Taken (t₁) = 8 s

Time Taken (t₂) = 4 s

Acceleration (a) = ?

Final Velocity (v) = ?

Distance to be found (s₂) = ?

STEP #1: First find a. To find a we can use the 2nd Kinematic equation of motion i.e.

s₁ = ut₁ + (1/2)at²

120 = 0 x 8 + (1/2) x a x 8²

120 = 0 + (1/2) x a x 64

120 = 32a

a = 120/32

a = 3.75 m/s²

STEP #2: Now we can find v by 1st Kinematic equation of motion i.e.

v = u + at₂

v = 0 + 3.75 x 4

v = 15 m

STEP #3: Now we can find s₂ by 3rd Kinematic equation of motion i.e.

v² = u² + 2as₂

15² = 0² + 2 x 3.75 x s₂

225 = 0 + 7.5s₂

7.5s₂ = 225

s₂ = 225/7.5

s₂ = 30 m

Thus, the distance travelled by body in 4 s is 30 m.