Question

A particle moves by constant acceleration from initial position. If the distance travel by particle in last second is $\frac{7}{16}$ of total distance travel by particle. Then, time of motion is

• A. $4s$
• B. $4/7s$
• C. $7s$
• D. None of these

Answer 1

The correct option is A $4s$

The distance travel by particle in $t$ seconds accelerated with $f$ is
$S=0+\frac{1}{2}f{t}^2\Rightarrow S=\frac{1}{2}f{t}^2$
According to question,
$S_t=\frac{7}{16}S$
$\Rightarrow 0+\frac{1}{2}f(2t-1)=\frac{7}{16}\left(\frac{1}{2}f{t}^2\right)$
$\Rightarrow 7t^2-32t+16=0$
$\Rightarrow (t-4)(7t-4)=0$
$\Rightarrow t=4,4/7$
But $t=4/7<1$
Hence, $t=4s$.