First slide

Rectilinear Motion

Question

A particle is moving in a straight line. and passes through a point o with a velocity of 6 ms^-1.The particle moves with a constant retardation of 2ms^-2 for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O?

Moderate

A

3s
B

8s
C

Never
D

4s

Solution

Let the particle moves toward right with velocity 6 m/s. Due to retardation, after time t₁, its velocity becomes zero.
$From v = u - a t \Rightarrow 0 = 6 - 2 \times t_{1}$
$\Rightarrow t_{1} = 3 \sec$
But retardation works on it for 4 sec. It means after reaching point A, direction of motion gets reversed and acceleration works on the particle for next one second.
$\begin{array}{l} S_{O A} = u t_{1} - \frac{1}{2} a t_{1}^{2} = 6 \times 3 - \frac{1}{2} (2) (3)^{2} = 18 - 9 = 9 m \\ S_{A B} = \frac{1}{2} \times 2 \times (1)^{2} = 1 m \\ ∴ S_{B C} = S_{0 A} - S_{A B} = 9 - 1 = 8 m \end{array}$
Now velocity of the particle at point B in return journey
$v = 0 + 2 \times 1 = 2 m / s$
In return journey from B to C, particle moves with constant velocity 2 m/s to cover the distance 8 m.
$Time taken = \frac{Distance}{Velocity} = \frac{8}{2} = 4 s$
Total time taken by particle to return at point 0 is
$T = t_{0 A} + t_{A B} + t_{B C} = 3 + 1 + 4 = 8 s .$

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