A particle moving in a straight line covers half the distance with a speed of 3 m/s. The other half of the distance is covered in two equal time intervals with uniform speeds of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is

Question

Toppr Admin · Accepted Answer

VavgTDTT-xx2-xD7-3-x429-x4215-116-136-130-130-5-6180-18039-4-61m-s

A particle moving in a straight line covers half the distance with a speed of 3 $m s^{- 1}$ . The other half of the distance is covered in two equal time intervals with speeds of 4.5 $m s^{- 1}$ and 7.5 $m s^{- 1}$ , respectively. Find the average speed of the particle during this motion.

$V_{a v g} \frac{T D}{T T} = \frac{x}{\frac{x}{2 \times 3} + \frac{x}{4} \frac{2}{9} + \frac{x}{4} \frac{2}{15}} = \frac{1}{\frac{1}{6} + \frac{1}{36} + \frac{1}{30}} = \frac{1}{\frac{30 + 5 + 6}{180}} = \frac{180}{39} = 4.61 m / s$

A particle moving in a straight line covers half the distance with a speed of 3 ms−1. The other half of the distance is covered in two equal time intervals with speeds of 4.5 ms−1 and 7.5 ms−1, respectively. Find the average speed of the particle during this motion.

VavgTDTT=xx2×3+x429+x4215=116+136+130=130+5+6180=18039=4.61m/s

A particle moving in a straight line covers half the distance with a speed of 3 $m s^{- 1}$ . The other half of the distance is covered in two equal time intervals with speeds of 4.5 $m s^{- 1}$ and 7.5 $m s^{- 1}$ , respectively. Find the average speed of the particle during this motion.

$V_{a v g} \frac{T D}{T T} = \frac{x}{\frac{x}{2 \times 3} + \frac{x}{4} \frac{2}{9} + \frac{x}{4} \frac{2}{15}} = \frac{1}{\frac{1}{6} + \frac{1}{36} + \frac{1}{30}} = \frac{1}{\frac{30 + 5 + 6}{180}} = \frac{180}{39} = 4.61 m / s$