Question

A particle starts to move in a straight line from a point with velocity $10m{s}^{-1}$ and acceleration $-2.0m{s}^{-2}$. Find position and velocity of the particle at t = 10 s

• A. S = 0 m, v = 10 m/s
• B. S = 0 m, v = -10 m/s
• C. S = 50 m, v = 10 m/s
• D. S = 0 m, v = 0 m/s

Answer 1

The correct option is B S = 0 m, v = -10 m/s

Displacement at t = 10 s is

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

$=10\times 10+\frac{1}{2}\times (-2.0)\times (10{)}^2$

$=100-100=0\left(zero\right)$

i.e., after 10 s, the particle will come back to the starting point.

Velocity at t = 10 s is

$v=u+at$

$v=10+(-2)\ast 10$

$=-10m{s}^{-1}$.

i.e., velocity is $10m{s}^{-1}$ opposite to the direction of initial starting velocity.