A car moves from X to Y with a uniform speed Vu and returns to Y with a uniform speed vd. The average speed for this round trip is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Question

A car moves from X to Y with a uniform speed Vu and returns to Y with a uniform speed vd. The average speed for this round trip is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is (A) (2 vd vu/vd + vu) (B)  √vuvd (C)  (vdvu/vd+vu) (D)  (vu+vd/2)

Tardigrade Admin · Accepted Answer

Average speed of a body in a given time interval is defined as the ratio of distance travelled to the time taken.
Average speed

= \frac{Distance travelled}{Time taken}

Let

t_{1}

and

t_{2}

be times taken by the car to go from

X

to

Y

and then from

Y

to

X

respectively.
Then,

t_{1} + t_{2} = \frac{X Y}{v _{u}} + \frac{X Y}{v _{d}}

= X Y (\frac{v _{u} + v _{d}}{v _{u} v _{d}})

Total distance travelled

= X Y + X Y = 2 X Y

Therefore, average speed of the car for this round trip is

v_{a v} = \frac{2 X Y}{X Y ( \frac{v _{u} + v _{d}}{v _{u} v _{d}} )}

or

v_{a v} = \frac{2 v _{u} v _{d}}{v _{u} + v _{d}}

Q. A car moves from $X$ to $Y$ with a uniform speed $V_{u}$ and returns to $Y$ with a uniform speed $v_{d}$ . The average speed for this round trip is $T$ . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

$\frac{2 v _{d} v _{u}}{v _{d} + v _{u}}$

$v_{u} v_{d}$

$\frac{v _{d} v _{u}}{v _{d} + v _{u}}$

$\frac{v _{u} + v _{d}}{2}$

Q. A car moves from X to Y with a uniform speed Vu​ and returns to Y with a uniform speed vd​. The average speed for this round trip is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

vd​+vu​2vd​vu​​

vu​vd​​

vd​+vu​vd​vu​​

2vu​+vd​​

Q. A car moves from $X$ to $Y$ with a uniform speed $V_{u}$ and returns to $Y$ with a uniform speed $v_{d}$ . The average speed for this round trip is $T$ . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

$\frac{2 v _{d} v _{u}}{v _{d} + v _{u}}$

$v_{u} v_{d}$

$\frac{v _{d} v _{u}}{v _{d} + v _{u}}$

$\frac{v _{u} + v _{d}}{2}$