Question

A particle moving along a straight line with constant acceleration is having initial and final velocity of 5 metre per second and 15 metre per second respectively in a time interval of 5 second find the distance travelled by the particle and the acceleration of the particle is the particle continues with the same acceleration find the distance covered by the particle in the eighth second of its motion​

Answer 1

s=1÷2+ut square formula should be used

Answer 2

Given,

u = 5m/s

v = 15m/s

t = 5s

CASE - 1

From the given data,

using first kinematic equation

$v = u + at$

15 = 5 + 5a

a = 2m/s^2

also, using second kinematic equation

$s = ut + \frac{a{t}^{2} }{2}$

s = 25 + (2 × 25)/2

s = 50m

CASE - 2

During eighth second,

$v = u + at$

15 = 5 + 8a

a = 10/8 m/s^2

$s = ut + \frac{a{t}^{2} }{2}$

s = 25 + (10×25)/16

s = 25 + 125/8

s = 325/8 m