# Kinematic Problem: Car Chase

• rachael123456
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.

## 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?

Write kinematic equations for car a and car b separately.
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.

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.

## 1. What is a kinematic problem?

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

## 2. What is a car chase kinematic problem?

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.

## 3. How do you solve a car chase kinematic problem?

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.

## 4. What factors can affect the outcome of a car chase kinematic problem?

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.

## 5. How is a car chase kinematic problem similar to a real-life car chase?

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.