A particle moving in a straight line covers half the distance with speed 6 m/s. The other half is covered in two equal time intervals with speeds 9 m/s and 15 m/s respectively. The average speed of the particle during the motion is: (a) 9.2 m/s (b) 8.8 m/s (c) 10 m/s (d) 8 m/s

Answer: (d). Let each half be d. First half time = d/6. Second half: equal times t at 9 and 15 m/s, so 9t + 15t = d ⇒ t = d/24, time = 2t = d/12. vavg = (2d)/(d/6 + d/12) = 2d/d/4 = 8 m/s So the answer is (d) 8 m/s.

A particle moving in a straight line covers half the distance with… | JEE Main 2024 (9 Apr, Shift 1)

Q7JEE Main 2024 · 9 APR · SHIFT 1Medium

Distance, Displacement & Average Speed

A particle moving in a straight line covers half the distance with speed $6\text{ m/s}$ . The other half is covered in two equal time intervals with speeds $9\text{ m/s}$ and $15\text{ m/s}$ respectively. The average speed of the particle during the motion is:

Show answer and solution

Answer: (d)

Let each half be $d$ . First half time $= \frac{d}{6}$ .
Second half: equal times $t$ at 9 and 15 m/s, so $9t + 15t = d \Rightarrow t = \frac{d}{24}$ , time $= 2t = \frac{d}{12}$ .
$v_{avg} = \frac{2d}{d/6 + d/12} = \frac{2d}{d/4} = 8$ m/s
So the answer is (d) 8 m/s.

Report a mistake

Distance, Displacement & Average Speed question from JEE Main 2024 (9 Apr, Shift 1)

More questions on Distance, Displacement & Average Speed