First slide

Position Path Length (Distance) And Displacement

Question

The displacement of a particle is given by $x = {(t - 2)}^{2},$ where x is in metre and t in second. The distance covered by the particle in first 4 seconds is

Moderate

A

4 m
B

8 m
C

12 m
D

16 m

Solution

Given, $x = {(t - 2)}^{2}$
Velocity,    $v = \frac{d x}{d t} = \frac{d}{d t} (t - 2)^{2} = 2 (t - 2) {ms}^{- 1}$
Acceleration,    $a = \frac{d v}{d t} = \frac{d}{d t} [2 (t - 2)] = 2 (1 - 0) = 2 {ms}^{- 2}$
When    $t = 0; v = - 4 {ms}^{- 1}, t = 2 s; v = 0 {ms}^{- 1}$
$t = 4 s; v = 4 m s^{- 1}$
v-t graph is shown in figure below.
Distance travelled = Area under the v-t graph
   $= ∣ Area O A C ∣ + Area A B D$
$= |\frac{4 \times 2}{2}| + \frac{1}{2} \times 2 \times 4 = 8 m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App