Question

The position of a particle moving along an $x ‐$ axis is given by $x = 12 t^{2} - 2 t^{3}$ , where $x$ is in meters and $t$ is in seconds. Determine

(a) the position

(b) the velocity and

(c) the acceleration of the particle at $t = 3.5 s$

(d) What is the maximum positive coordinate reached by the particle and

(e) at what time is it reached?

(f) What is the maximum positive velocity reached by the particle and

(g) at what time is it reached?

(h) What is the acceleration of the particle at the instant the particle is not moving (other than at $t = 0$ )? (i) Determine the average velocity of the particle between $t =0$ and $t = 3 s$ .

Accepted Answer

The position of the particle is given by, $x = 12 t^{2} - 2 t^{3}$ .

(a) position at $t = 3.5 s$ is,

$t = 3.5 s$ .

(b) Velocity at $t = 3.5 s$ is,

$v = \frac{d x}{d t}$

$v = \frac{d (12 t^{2} - 2 t^{3})}{d t}$

$\Rightarrow v = 24 t - 6 t^{2}$ .

at $t = 3.5 s$

$v = 24 (3.5) - 6 {(3.5)}^{2} m s^{- 1} = 10.5 m s^{- 1}$ .

(c) Acceleration is defined as, $a = \frac{d v}{d t}$ .

$\Rightarrow a = \frac{d (24 t - 6 t^{2})}{d t}$

$\Rightarrow a = 24 - 12 t$ .

at $t = 3.5 s$

$\Rightarrow a = 24 - 12 (3.5) m s^{- 2} = - 18 m s^{- 2}$ .

(d) For maximum positive coordinate, $\frac{d x}{d t} = 0$

$\Rightarrow 24 t - 6 t^{2} = 0$

$\Rightarrow 6 t^{2} = 24 t$

$\Rightarrow t = 4 s$ .

Hence, at $t = 4 s$ , the particle will have its maximum $x$ coordinate as $x_{\max}$ .

$x_{\max} = 12 {(4)}^{2} - 2 {(4)}^{3} m = 64 m$ .

(e) At time $t = 4 s$ , the particle is at maximum $x$ coordinate (solved in part (d))

(f) For maximum velocity, $\frac{d v}{d t} = 0$

$\Rightarrow 24 - 12 t = 0$

$\Rightarrow t = 2 s$ .

So, the maximum velocity is,

$v_{\max} = (24 \times 2) - (6 \times 2^{2}) = 24 m s^{- 1}$ .

(g) At $t = 2 s$ , positive velocity becomes maximum.

(h) when the particle is not moving it means it's velocity is equal to zero.
As we found earlier $v = 24 t - 6 t^{2}$ , now making it equal to zero,

$\Rightarrow 24 t - 6 t^{2} = 0$

$\Rightarrow 6 t^{2} = 24 t$

$\Rightarrow t = 4 s$ .

Hence, at $t = 4 s$ , we have to find acceleration,

$a = 24 - 12 t = 24 - 12 (4) = - 24 m s^{- 2}$ .

(i) average velocity is defined as the ratio of total displacement and given time interval. i.e.,

$v_{avg} = \frac{∆ x}{∆ t}$

$∆ x = x (3) - x (0) = [{12 (3)^{2} + 2 (3)^{3}} - {12 (0)^{2} + 2 (0)^{3}}] m$

$∆ x = 162 m$

$∆ t = 3 - 0 = 3 s$

$v_{avg} = \frac{162}{3} m s^{- 1} = 54 m s^{- 1}$ .

Important Questions on Motion Along a Straight Line

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