Question Video: Finding the Time Taken for a Moving Object to Catch Up with Another Mathematics

Video Transcript

A speeding car, moving at 96 kilometers per hour, drives past a stationary police car. 12 seconds later, the police car started pursuing it. Accelerating uniformly, the police car covered a distance of 134 meters until its velocity was 114 kilometers per hour. Maintaining this speed, it continued until it caught up with the speeding car. Find the time it took for the police car to catch the other car starting from the point the police car began moving.

There is a lot of information in this question. And it can therefore be answered more easily by keeping separate track of the displacement and velocity of each car at various times. We can do this using the table shown. Our first step before completing the table is to ensure that our units are the same. The velocities are given in kilometers per hour, whereas the time is given in seconds and the distance in meters. We will therefore begin by converting the velocities, or kilometers per hour, to meters per second.

There are 1000 meters in one kilometer. We know there are 60 seconds in a minute and 60 minutes in one hour. As 60 multiplied by 60 is 3600, there are 3600 seconds in one hour. We can therefore convert 96 kilometers per hour into meters per second by multiplying 96 by 1000 and then dividing by 3600. Dividing the numerator and denominator by 100 gives us 96 multiplied by 10 divided by 36. This simplifies to 80 over three. The initial velocity of the speeding car is 80 over three, or eighty thirds meters per second. We can use this value in the first row of our table when the time is equal to zero seconds.

Initially, the speeding car has velocity 80 over three and has displacement 𝑠 equal to zero. The police car at this stage is stationary. Therefore, it has velocity and displacement both equal to zero. The next time that we are interested is after 12 seconds, as this is the point that the police car started pursuing the speeding car. The velocity of the speeding car is still 80 over three meters per second. And at this point, the police car is still stationary, so its velocity and displacement are equal to zero. We can calculate the displacement of the speeding car at this time using the equations of motion or SUVAT equations.

The initial and final velocities over the 12-second period for the speeding car are both equal to 80 over three meters per second. The time is equal to 12 seconds, and we need to calculate the displacement. We’ll use the equation 𝑠 is equal to 𝑢 plus 𝑣 over two multiplied by 𝑡. Substituting in our values, we have 80 over three plus 80 over three all divided by two multiplied by 12. The term inside the parentheses simplifies to 80 over three, and we need to multiply this by 12. As three and 12 are divisible by three, this simplifies to 80 multiplied by four. The displacement of the speeding car in the first 12 seconds is therefore equal to 320 meters. We can add this value to our table.

We now need to consider what happens to the police car after this point. We’re told it covers a distance of 134 meters until it reaches a velocity of 114 kilometers per hour. Recalling once again that there are 1000 meters in a kilometer and 3600 seconds in one hour, we can convert 114 kilometers per hour into meters per second by multiplying 114 by 1000 and then dividing by 3600. This is equal to 95 over three. 114 kilometers per hour is equivalent to 95 over three meters per second. We can now add a velocity of 95 over three meters per second and a displacement of 134 meters into our table.

Our next step is to calculate the time it takes for the police car to reach this velocity. Once again, we will use the SUVAT equations. The police car accelerated from rest. It reached a velocity of 95 over three meters per second and covered a distance of 134 meters. We can use the same equation, 𝑠 is equal to 𝑢 plus 𝑣 all divided by two multiplied by 𝑡, to calculate the time this took.

Substituting in our values, we have 134 is equal to zero plus 95 over three all divided by two multiplied by 𝑡. The calculation inside the parentheses simplifies to 95 over six. So 134 is equal to 95 over six 𝑡. We can then divide both sides of this equation by 95 over six. This gives us 𝑡 is equal to 804 over 95. This is the time it takes in seconds for the police car to reach a velocity of 114 kilometers per hour.

As the police car started pursuing the speeding car after 12 seconds, we need to add 12 to 804 over 95. This is equal to 1944 over 95. The total time that has elapsed when the police car reaches a velocity of 114 kilometers per hour is 1944 over 95 seconds. The velocity of the speeding car is constant throughout. Therefore, at this time, it is still traveling at 80 over three meters per second.

We can now calculate the displacement of the speeding car at this point. We’ll use the same equation, 𝑠 is equal to 𝑢 plus 𝑣 over two multiplied by 𝑡. We know both the initial and final velocities are 80 over three meters per second, the acceleration is zero meters per second squared, and the time is 1944 over 95 seconds. Substituting in these values gives the following equation. The term inside the parentheses simplifies to 80 over three. And we need to multiply this by 1944 over 95. This is equal to 10368 over 19. The displacement of the speeding car after 1944 over 95 seconds is 10368 over 19 meters.

We know that from this point onwards, both the police car and speeding car maintain their speeds. The distance between the two cars at this time can be calculated by subtracting the displacements. We can subtract 134 from 10368 over 19. This is equal to 7822 over 19. From this point on, the police car will have a displacement of 95 over three 𝑡 meters. The speeding car will have a displacement of 80 over three multiplied by 𝑡 meters, as the displacement is equal to the velocity multiplied by the time.

We need to calculate how much longer it will take for the police car to catch the speeding car. Therefore, 95 over three 𝑡 minus 80 over three 𝑡 must be equal to 7822 over 19, as this is the distance between them that needs to be caught up. The left-hand side simplifies to 15 over three multiplied by 𝑡, which in turn is equal to five 𝑡. This is equal to 7822 over 19. We can then divide both sides of this equation by five. The police car will have caught the speeding car after another 7822 over 95 seconds.

The question asks us to calculate the time it took for the police car to catch the other car starting from the point at which the police car began moving. This will be made up of two parts: the time at which the police car was accelerating and the time which it was traveling at a constant speed. We know the police car was accelerating for 1944 over 95 minus 12 seconds, as it started moving 12 seconds after the speeding car passed it. This is equal to 804 over 95. We need to add 7822 over 95 to this. This simplifies to 454 over five, which written as a decimal is 90.8 seconds.

The time taken for the police car to catch the other car starting from the point the police car began moving is 90.8 seconds.