A car A is moving with a uniform velocity of 44m/s over takes another car B which is just moving from rest with an acceleration of 2m/s.after how much time do they meet again?

Question

Answer

Problem Statement

A car A is moving with a uniform velocity of 44m/s overtakes another car B which is just moving from rest with an acceleration of 2m/s. After how much time do they meet again?

Solution

Step 1: Find the distance covered by car A

Let us assume that the two cars meet again after time t. During this time, car A covers a certain distance d, which can be calculated as:

d = v*t

where v is the velocity of car A, which is 44m/s.

So, d = 44t

Step 2: Find the distance covered by car B

During the same time t, car B travels a distance which can be calculated using the equation:

d = ut + (1/2)at^2

where u is the initial velocity of car B, which is 0m/s, a is the acceleration of car B, which is 2m/s^2 and t is the time taken.

So, d = (1/2)*2*t^2

d = t^2

Step 3: Equate the distances

Since the two cars meet again, the distance covered by both of them must be equal. Therefore, we can equate the two distances calculated in step 1 and step 2:

44t = t^2

t^2 - 44t = 0

t(t - 44) = 0

Either t = 0 (which is not possible) or t = 44 seconds.

Step 4: Interpret the result

Therefore, the two cars will meet again after 44 seconds.

Answer

22 minutes

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