Question

A particle moving along a straight line with constant acceleration is having initial and final velocity as $5m/s$ and $15m/s$ respectively in a time interval of $5s$. Find the distance travelled by the particle and the acceleration of the particle. If the particle continues with same acceleration, find the distance covered by the particle in the $8th$ second of its motion (direction of motion remains same).

Answer 1

Initial velocity, $u=5m/s$

Final velocity, $v=15m/s$

Time interval, $t=5s$

$a$ is the constant acceleration with which the particle is moving.

$s$ is the distance covered by the particle in the given time interval.

$v=u+at$

$15=5+a\times 5$

$a=2m/s^2$

Using the third equation of motion,

$v^2=u^2+2as$

$225=25+2\times 2\times s$

$s=\frac{200}{4}=50m$

Using the second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

Distance covered in $8s$ =$5\times 8+\frac{1}{2}\times 2\times (8{)}^2=40+64=104m$

Distance covered in $7s$ =$5\times 7+\frac{1}{2}\times 2\times (7{)}^2=35+49=84m$

The distance covered in $8th$ second of its motion = Distance covered in $8s$ - Distance covered in $7s$

= $104m-84m$= $20m$