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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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In summary, the two cars, car a starting from rest with an acceleration of 4.5m/s² and car b with a constant velocity of 25m/s, start at different distances but eventually meet when car a catches up to car b. By drawing velocity-time graphs and finding the areas between them, it can be determined that car b will catch up to car a after approximately 17 seconds, or 14 seconds in real time. Additionally, kinematic equations can also be used to solve for the time it takes for car b to catch up to car a.

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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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rachael123456 said:## 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 traveling 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.

A kinematic problem is a type of physics problem that deals with the motion of objects without considering the forces that cause the motion.

A car chase kinematic problem involves analyzing the motion of multiple cars during a chase scene, taking into account their initial positions, velocities, and accelerations.

To solve a car chase kinematic problem, you can use equations of motion, such as the kinematic equations, to calculate the position, velocity, and acceleration of each car at different points in time.

The outcome of a car chase kinematic problem can be affected by factors such as the acceleration and braking abilities of the cars, the road conditions, and any obstacles or traffic in the way.

A car chase kinematic problem is similar to a real-life car chase in that both involve analyzing the motion of multiple cars and taking into account their speeds, positions, and accelerations. However, in a real-life car chase, there are also external factors such as driver skill and strategy that can affect the outcome.

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