Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

Position (in

m

) of a particle moving on a straight line varies with time (in

s

) as

x = \frac{t^{3}}{3} - 3 t^{2} + 8 t + 4

(m)

. Consider the motion of the particle from

t = 0

to

t = 5 s

. Here,

s_{1}

is the total distance travelled and

s_{2}

is the distance travelled during retardation. Find

\frac{s_{1}}{s_{2}}

.

Accepted Answer

We know that position of a particle,

$x = \frac{t^{3}}{3} - 3 t^{2} + 8 t + 4 . . . (1)$

On differentiating the equation $(1)$ with respect to time, we get velocity as function of time.

$\Rightarrow v = \frac{d x}{d t} = \frac{1}{3} (3 t^{2}) - 3 (2 t) + 8 (1) + 0$

$\Rightarrow v = t^{2} - 6 t + 8 . . . (2)$

$\Rightarrow v = (t - 2) (t - 4)$

On putting velocity as zero,

$(t - 2) (t - 4) = 0$

$\Rightarrow t = 2 s$ and $s$ .

At $t = 2 s$ and $t = 4 s$ velocity changes its direction.

On putting $t = 0$ in the equation $(1)$ , we get,

$\Rightarrow x = 4 m$ .

On putting $t = 2 s$ in the equation $(1)$ , we get,

$\Rightarrow x = \frac{2^{3}}{3} - 3 {(2)}^{2} + 8 (2) + 4$

$\Rightarrow x = \frac{32}{3} m$ .

On putting $t = 4 s$ in equation $(1)$ , we get,

$\Rightarrow x = \frac{4^{3}}{3} - 3 {(4)}^{2} + 8 (3) + 4$

$\Rightarrow x = \frac{28}{3} m$ .

On putting $t = 5 s$ in the equation $(1)$ , we get,

$\Rightarrow x = \frac{5^{3}}{3} - 3 {(5)}^{2} + 8 (5) + 4$

$\Rightarrow x = \frac{32}{3} m$ .

On differentiating the equation $(2)$ with respect to time, we get acceleration as a function of time.

$\Rightarrow a = \frac{d v}{d t} = 2 t - 6$

$\Rightarrow a = 2 (t - 3)$ , here we observe that acceleration is negative (retardation) between $t = 0$ to $t = 3 s$ and after that it is positive.

Hence, from the diagram, the total distance travelled $s_{1}$

$= (\frac{32}{3} - 4) + 2 (\frac{32}{3} - \frac{28}{3})$

$\Rightarrow s_{1} = \frac{28}{3} m$ .

On putting $t = 3 s$ in the equation $(1)$ , we get,

$\Rightarrow x = \frac{3^{3}}{3} - 3 {(3)}^{2} + 8 (3) + 4$

$\Rightarrow x = 10 m$ .

Distance travelled during retardation (between $t = 0$ to $t = 3 s$ ) is,

$\Rightarrow s_{2} = (\frac{32}{3} - 4) + (\frac{32}{3} - 10)$

$\Rightarrow s_{2} = \frac{20}{3} + \frac{2}{3}$

$\Rightarrow s_{2} = \frac{22}{3} m$ .

We have to find $\frac{s_{1}}{s_{2}} = \frac{\frac{28}{3}}{\frac{22}{3}}$

$\Rightarrow \frac{s_{1}}{s_{2}} = \frac{14}{11}$ .

Important Questions on Kinematics

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Position (in m) of a particle moving on a straight line varies with time (in s) as x=t33-3t2+8t+4 m. Consider the motion of the particle from t=0 to t=5 s. Here, s1 is the total distance travelled and s2 is the distance travelled during retardation. Find s1s2.

Important Questions on Kinematics

Learn and Prepare for any exam you want !