Question

A body moving with uniform acceleration covers a distance of 14m in first 2 second and 88m in next 4 second. How much distance is travelled by it in 5 th second

Answer 1

Let u is initial velocity and a is uniform acceleration of body.

case 1 : s = 14m , t = 2

using formula, s = ut + 1/2 at²

or, 14 = u(2) + 1/2 a(2)²

or, 14 = 2u + 2a

or, u + a = 7 .......(1)

velocity after t = 2s , v = u + at ⇒v = u + a(2) = u + 2a......(2)

case 2 : s = 88m , t = 4s, v = u + 2a

using , s = vt + 1/2 at²

or, 88 = (u + 2a)(4) + 1/2a(4)²

or, 88 = 4u + 8a + 8a

or, 88 = 4u + 16a

or, u + 4a = 11.......(3)

from equations (1) and (3) we get,

u + 4a - u - a = 11 - 7

or, 3a = 4

or, a = 4/3 m/s²

so, u = 7 - 4/3 = 17/3 m/s

so, distance travelled in 5th second = u + a(2n - 1)/2

= 17/3 + (4/3)(2 × 5 - 1)/2

= 17/3 + 4/3 × 9/2

= 17/3 + 18/3

= 35/3 m

= 11.667 m

Answer 2

Answer:

Thus, the body travel 125m in 5th seconds

Explanation:

The body is moving under uniform acceleration.

so,

Let the velocity for 2nd and 4th second be $v_{2}$ and $v_{4}$ is -

V{2} = 14 m/s

V{4} = 88 m/s

Now,

The rate of change of velocity = $\dfrac{v_{4}-v_{2}}{time}$

                                                    = $\dfrac{88-14}{2}$

                                                   = $\dfrac{74}{2}$

                                                    = 37 $ms^{-2}$

Then, the rate of change in velocity is equal to the uniform acceleration

therefore, let 'a' be the acceleration

then, a = 37 $ms^{-2}$

Now, we have to find the distance traveled in 5th second

let time t = 5 and velocity for 5th second be $v_{5}$

According to the formula of velocity,

v = u + at ----------------(i)

where, v = $v_{5}$

           u = velocity of 2nd second,$v_{2}$

           a = acceleration

then,

          $v_{5}$  = $v_{2}$  + at

          $v_{5}$ = 14+(37*(5-2))

          $v_{5}$ = 125 m/s