Question

Two cars are traveling from the same point. Car A has a velocity of 12.5 m/s while car B has zero velocity but an acceleration of 3.5 m/s 2 . How far apart will the cars be after 12.0 seconds? What are the final velocities for Cars A and B?

Answer 1

The distance of separation of the two cars after 12.0 seconds is 102 m

Final velocity, v of A = 12.5 m/s

Final velocity, v of B = 45 m/s

What are the distance apart of the two car after 12 seconds?

The distance of separation of the two cars after 12.0 seconds is calculated as follows:

• The distance travelled by A = velocity * time

Distance travelled by A = 12.5 * 12.0

Distance travelled by A = 150 m

Distance travelled by B is calculated as follows:

• Distance, s = ut + 1/2at²

u = 0, t = 12.0, a = 3.5 m/s²

Distance travelled by B = 0 * 12.0 + 1/2 * 3.5 * 12²

Distance travelled by B = 252 m

The distance of separation of the two cars after 12.0 seconds = 252 - 150 The distance of separation of the two cars after 12.0 seconds = 102 m

Final velocity, v of A = 12.5 m/s

Final velocity, v of B is calculated as follow;

v = u + at

v = 0 + 3.5* 12

Final velocity of B, v = 45 m/s

In conclusion, the distance of separation and final velocities are determined from the given values of initial velocity and acceleration.

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