Question

A dancer demonstrating dance steps along a straight line. The position-time graph is given below.

Find the average velocity of the dancer during the time interval between t $=$ 4.5 s to t $=$ 9 s

• A. $1m{s}^{-1}$
• B. $-1.33m{s}^{-1}$
• C. $2.75m{s}^{-1}$
• D. $-0.89m{s}^{-1}$

Answer 1

The correct option is D $-0.89m{s}^{-1}$

Displacement = final position - initial position

From graph final position is zero at $9s$.

From graph initial position is $4m$ at $4.5s$.

Here, displacement $=-4m$
Time taken,

$t=9-4.5$

$\Rightarrow t=4.5s$

Average velocity$\left(V_{avg}\right)$

$V_{avg}=\frac{displacement}{time}$
$\Rightarrow V_{avg}=\frac{-4}{4.5}$

$\Rightarrow V_{avg}=-0.89m{s}^{-1}$