Ncert Exercises & Solutions

A body of mass $2~\text{kg}$ travels according to the law $x \left( t \right) = pt + qt^2+ rt^3$ where, $p = 3 ~\text{ms }^{−1 },$ $q = 4 ~\text{ms }^{−2}$ and $r = 5 ~\text{ms }^{−3}$ . The force acting on the body at $t = 2 ~\text{s }$ is

136 N

$136~\text{N}$
2.

134 N

$134~\text{N}$
3.

158 N

$158~\text{N}$
4.

68 N

$68~\text{N}$

(a) Hint: Using the equation of motion of the particle, we can find the acceleration of the particle.

Step 1: Find the acceleration and force acting on the particle.

Given, mass = 2 kg

\begin{array}{l} x (t) = pt + {qt}^{2} + {rt}^{3} \\ v = \frac{dx}{dt} = p + 2 qt + 3 {rt}^{2} \\ a = \frac{dv}{dt} = 0 + 2 q + 6 rt \\ at t = 2 s; a = 2 q + 6 \times 2 \times r \\ = 2 q + 12 r \\ = 2 \times 4 + 12 \times 5 \\ = 8 + 60 = 68 m / s \\ Force = F = ma \\ = 2 \times 68 = 136 N \end{array}

$\begin{array}{l} x (t) = pt + {qt}^{2} + {rt}^{3} \\ v = \frac{dx}{dt} = p + 2 qt + 3 {rt}^{2} \\ a = \frac{dv}{dt} = 0 + 2 q + 6 rt \\ at t = 2 s; a = 2 q + 6 \times 2 \times r \\ = 2 q + 12 r \\ = 2 \times 4 + 12 \times 5 \\ = 8 + 60 = 68 m / s \\ Force = F = ma \\ = 2 \times 68 = 136 N \end{array}$

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