Question

A body moving along a straight line travels one third of the total distance with a velocity 4 m/s. The remaining distance is covered with velocity 2 m/s for some time t0 and with 4 m/s for the remaining time. If the average velocity is 3 m/s, then the time for which the body moves with velocity 4 m/s in second stretch is
3/2t
t
2t
t/2​

Answer 1

2t is the correct answer .

Answer 2

Answer:

The correct answer is Option(d) $\frac{t}{2}$

Explanation:

Given:

• One third of the total distance with a velocity = 4m/s
• The remaining distance is covered with velocity = 2 m/s at t₀
• The distance covered after t₀ with the velocity = 4m/s
• If the average velocity  = 3m/s
• t₀ is the initial time

To Find:

The time for which the body moves with velocity 4m/s in second stretch.

Solution:

$Required time = k t_{0}$    ...............................(1)

To find the value of k, find the average velocity:

$Average velocity = \frac{s}{t_{1} + t_{0} +t_{0} k }$    .......................(2)

Find the value of  t₀ and t₁ and substitute in equation (2) we get

$t_{1} = \frac{(\frac{s}{3}) }{4}\\ \\ t_{1} = \frac{s}{12}$ ....................................(3)

$\frac{2s}{3} = 2t_{0} + 4 (kt_{0})\\ \\ \frac{2s}{3} = t_{0} ( 2+ 4k)\\ \\ t_{o} = \frac{2s}{3( 2+ 4k)}$.........................(4)

Substitute equation (3) and (4) in equation (2) we get:

$Average velocity = \frac{s}{(\frac{s}{12} )+\frac{2s(1+k)}{(5+6k)} }$

$Average velocity = \frac{6(2+4k)}{(5+6k)}$

$V_{avg} (5+6k) = 12 +24k$

$k = \frac{1}{2}$   .................................(5)

Substitute the value of k in equation (1) we get:

$Required time = kt_{0} or kt\\\\ Required time = \frac{t_{0} }{2} or \frac{t}{2}$

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