Question

28. A particle start moving from rest state along a
straight line under the action of a constant force
and travel distance x in first 5 seconds. The
distance travelled by it in next five seconds will be
(1) x
(2) 2x
(3) 3x
(4) 4x​

Answer 1

Since the particle is acted upon a constant force, it has constant acceleration if the mass of the particle remains unchanged.

The distance travelled by the particle in first 5 seconds is given by second equation of motion as,

$\longrightarrow\sf{x=0\times5+\dfrac{1}{2}\times a\times5^2}$

$\longrightarrow\sf{x=12.5\,a}$

The distance travelled by the particle in first 10 seconds is, by second equation of motion,

$\longrightarrow\sf{x'=0\times10+\dfrac{1}{2}\times a\times10^2}$

$\longrightarrow\sf{x'=50\,a}$

$\longrightarrow\sf{x'=4\times12.5\,a}$

$\longrightarrow\sf{x'=4x}$

Hence the distance travelled by the particle in the final 5 seconds is,

$\longrightarrow\sf{s=x'-x}$

$\longrightarrow\sf{s=4x-x}$

$\longrightarrow\underline{\underline{\sf{s=3x}}}$

Hence (3) is the answer.