Question bank/JEE Main/Physics

Consider a particle initially moving with a velocity 7 m/s
starts decelerating at a constant rate of$2m/s^{2}$.
The distance travelled in the fourth second is$\frac{x}{10}m$. Find
the value of x.

A5B6C7D8

Show answer and solution▾

Here particle is decelerating so velocity of particle becomes
after t=3.5 sec. the particle has a turning point at t=3.5 sec.
V=U+at
$0 = 7 - 2t$
$t = 3.5\sec$
$y_{2} = 7 \times 3.5 - \frac{1}{2} \times 2 \times (3.5)^{2} = \frac{49}{4}m$
$y_{2} - x_{1} = \frac{49}{4} - 12 = \frac{1}{4}m$.
Due to symmetry displacement of the particle at t=3 and t=4sec. is same.
so total distance traveled in the fourth sec
is$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}m$

Source: Grass-G / P-22 Physcis Paper 19 Dec Eng..docx