Question

a body with an initial velocity of 3ms^-1 moves with an acceleration of 2ms^-2 then the distance travelled in 4th seconds is

Answer 1

s=ut+1/2at^2
= 2*4+1/2*2*4^2
= 8+16
=24m

Answer 2

Concept:

Numerous dimensional motion issues that take into account the motion of an object with constant acceleration can be solved using the kinematic equations. The second kinematical equation for accelerated motion in one dimension is given by the formula, s = ut + 1/2 at²

Given:

Initial velocity of body, u = 3m/s

Acceleration of body, a = 2m/s

Time, t = 4s

Find:

We need to determine the distance travelled by the body in the 4th second, d.

Solution:

Numerous dimensional motion issues that take into account the motion of an object with constant acceleration can be solved using the kinematic equations.

Here, we are going to use the second kinematical equation given as,

s = ut + 1/2 at²  where s = displacement, u = intial velocity, t = time period, a = acceleration

For time period, t = 4, second kinematical equation becomes,

s = 3×4 + 1/2×2×4²

s = 12 + 16

s = 28m

This is the total distance travelled by a body in 4 seconds.

For the time period, t = 3, the second kinematical equation becomes,

s = 3×3 + 1/2×2×3²

s = 9 + 9

s = 18m

This is the total distance travelled by a body in 3 seconds.

To calculate the distance travelled by a body in the 4th second, we need to calculate the difference between the distance travelled by a body in 4 seconds and that travelled in 3 seconds.

Distance travelled in 4th-second = total distance travelled by a body in 4 seconds - total distance travelled by a body in 3 seconds

d = 28 - 18

d = 10m

Thus, the distance travelled in the 4th second is 10m.

#SPJ3