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

$- 1.33 m s^{- 1}$
$2.75 m s^{- 1}$
$1 m s^{- 1}$
$- 0.89 m s^{- 1}$

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

$- 1.33 m s^{- 1}$
$2.75 m s^{- 1}$
$1 m s^{- 1}$
$- 0.89 m s^{- 1}$

Toppr Admin · Accepted Answer

Displacement - final position - initial positionFrom graph final position is zero at 9-xA0-s-From graph initial position is 4-xA0-m at 4-5-xA0-s-Here- displacement -x2212-4-xA0-mTime taken-xA0-t-9-x2212-4-5-x21D2-t-4-5-xA0-sAverage velocity-Vavg-Vavg-displacementtime-x21D2-Vavg-x2212-44-5-x21D2-Vavg-x2212-0-89ms-x2212-1

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−1.33ms−12.75ms−11ms−1−0.89ms−1

Displacement = final position - initial positionFrom graph final position is zero at 9 s.From graph initial position is 4 m at 4.5 s.Here, displacement =−4 mTime taken, t=9−4.5⇒t=4.5 sAverage velocity(Vavg)Vavg=displacementtime⇒Vavg=−44.5⇒Vavg=−0.89ms−1

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

$- 1.33 m s^{- 1}$
$2.75 m s^{- 1}$
$1 m s^{- 1}$
$- 0.89 m s^{- 1}$