Question Video: Finding the Stopping Distance of a Uniformly Decelerating Car after Braking | Nagwa Question Video: Finding the Stopping Distance of a Uniformly Decelerating Car after Braking | Nagwa

# Question Video: Finding the Stopping Distance of a Uniformly Decelerating Car after Braking Mathematics • Second Year of Secondary School

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A man was driving his car in a straight line at 78 km/h when he pressed the brakes. If the carโs velocity decreased at a constant rate until it stopped completely over a period of 15 seconds, determine the stopping distance of the car.

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### Video Transcript

A man was driving his car in a straight line at 78 kilometers per hour when he pressed the brakes. If the carโs velocity decreased at a constant rate until it stopped completely over a period of 15 seconds, determine the stopping distance of the car.

In this question, we will begin by modeling the situation. We are told that the car is initially traveling at 78 kilometers per hour. After pressing the brakes, the car comes to a complete stop. We are told that this takes a period of 15 seconds and to try and to calculate the stopping distance. We can answer the question using the equations of motion or SUVAT equations, where ๐  is the displacement. ๐ข and ๐ฃ are the initial and final velocities. ๐ is the acceleration. And ๐ก is the time. The standard units for these are meters, meters per second, meters per second, meters per second squared, and seconds, respectively.

Recalling that there are 1000 meters in one kilometer and 3600 seconds in one hour, our first step is to convert 78 kilometers per hour to meters per second. We can do this by multiplying 78 by 1000 over 3600. This is equal to 65 over three. The initial velocity of the car is therefore equal to 65 over three meters per second. And the final velocity is zero meters per second. We now have values in standard units for the initial and final velocities together with the time. We need to calculate the displacement, which we will call ๐ฅ meters.

One of the equations of motion states that ๐  is equal to ๐ข plus ๐ฃ divided by two multiplied by ๐ก. Substituting in our values, we have ๐ฅ is equal to 65 over three plus zero all divided by two multiplied by 15. This simplifies to 65 over six multiplied by 15, which in turn is equal to 325 over two, or 162.5. We can therefore conclude that the stopping distance of the car is 162.5 meters.

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