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## Homework Statement

Car a: starts moving from rest, acceleration= 4.5m/s², starting distance = 0m

car b : constant velocity of 25m/s, starting distance = 75m

How long will it take car b to catch up with car a?

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- Thread starter rachael123456
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Car a: starts moving from rest, acceleration= 4.5m/s², starting distance = 0m

car b : constant velocity of 25m/s, starting distance = 75m

How long will it take car b to catch up with car a?

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For car a:d=0.5*a*(t^2)

For car b:s=v*t+starting distance.

equate d and s.Get t

Here a=4.5(acceleration) v=25(velocity) t is time in seconds.

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PeterO

Homework Helper

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## Homework Statement

Car a: starts moving from rest, acceleration= 4.5m/s², starting distance = 0m

car b : constant velocity of 25m/s, starting distance = 75m

How long will it take car b to catch up with car a?

You can "see" the answer if you draw velocity time graphs of the two cars.

SInce Car b is 75 m in front, when you start looking at them, you might imagine that it started 3 seconds earlier than the Car a.

[25 m/s for 3 seconds will give you 75 m]

The v-t graph for car b will be a horizontal line at v = 25

The v-t graph for Car a will be a sloping line starting at t=3 [seconds start for (b) remember]

The gradient of the line will be 4.5.

Once you have both lines, you will see two areas.

Firstly below Car b, above Car a - a trapezium. That area represents how far in front Car b will be by the time Car a is also travelling at 25m/s

Secondly a triangular area above Car b / below Car a [it becomes a triangle when you draw a vertical line at time t].

This triangular area represents the extra distance covered by Car a as it catches up.

When those two areas are equal, Car a has caught Car b.

Note that I started this example 3 seconds before the problem, so if it works out t = 17 seconds, Car a actually catches up after 14 seconds. [I made up the value 17 - hope I didn't guess the right answer]

The equations suggested by Bhaskar are excellent too.

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