Question

Two cars X and Y start from two points separated by 75 m. Y which is ahead of X. starts from rest with acceleration of 10 m/s2 and X starts with uniform velocity of 40 m/s . They meet each other twice in their journey. Find the time gap between their meetings.

Answer 1

Answer:

The time gap between their meeting is 2 s

Explanation:

As we know that the distance between the two cars is initially given as

$d = 75 m$

now we know that

initial speed of Y = 0

acceleration of Y = 10 m/s/s

initial speed of X = 40 m/s

acceleration of X = 0

now by the concept of relative motion we have

$d = v_{rel} t + \frac{1}{2}a_{rel} t^2$

$75 = 40 t + \frac{1}{2}(-10) t^2$

$5t^2 - 40 t + 75 = 0$

$t^2 - 8 t + 15 = 0$

$(t - 3)(t - 5) = 0$

$t = 3 s , 5 s$

so the time interval between two positions where the two cars will meet is given as

$\Delta t = 5 - 3 = 2 s$

#Learn

Topic : Relative Motion

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