Question Video: Finding the Distance Covered by a Cyclist Where They Moved with Uniform Acceleration Then at Uniform Velocity | Nagwa Question Video: Finding the Distance Covered by a Cyclist Where They Moved with Uniform Acceleration Then at Uniform Velocity | Nagwa

# Question Video: Finding the Distance Covered by a Cyclist Where They Moved with Uniform Acceleration Then at Uniform Velocity Mathematics • Second Year of Secondary School

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A cyclist riding down a hill from rest was accelerating at a rate of 0.5 m/sĀ². By the time he reached the bottom of the hill, he was traveling at 1.5 m/s. He continued traveling at this speed for another 9.5 seconds. Determine the total distance š  that the cyclist covered.

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### Video Transcript

A cyclist riding down a hill from rest was accelerating at a rate of 0.5 meters per second squared. By the time he reached the bottom of the hill, he was traveling at 1.5 meters per second. He continued traveling at this speed for another 9.5 seconds. Determine the total distance š  that the cyclist covered.

In this question, we are told that a cyclist starting from rest rides down the hill. He accelerates at a rate of 0.5 meters per second squared. And by the time he reaches the bottom of the hill, he was traveling at 1.5 meters per second. As he started from rest, we know that the initial speed in this part of the journey is zero meters per second. Once the cyclist reached the bottom of the hill, we are told he continues to travel at the speed of 1.5 meters per second for 9.5 seconds. We are asked to calculate the total distance š  that he travels. We will split this into two parts: š  sub one, the journey on the hill, and š  sub two, the journey on the flat.

We can calculate the value of š  sub one using our equations of motion, or SUVAT equations, as the cyclist is traveling with uniform acceleration. As already mentioned, the initial velocity on this part of the journey is zero meters per second, and the final velocity is 1.5 meters per second. The acceleration is 0.5 meters per second squared. And we are trying to calculate the distance or displacement š  sub one. We will use the equation š£ squared is equal to š¢ squared plus two šš . Substituting in our values, we have 1.5 squared is equal to zero squared plus two multiplied by 0.5 multiplied by š  sub one. The left-hand side is equal to 2.25. And as two multiplied by 0.5 is one, this is equal to š  sub one. The distance traveled by the cyclist down the hill is 2.25 meters.

During the second part of the journey, the cyclist is traveling at a constant speed of 1.5 meters per second. This means that his acceleration is equal to zero. And the speed is equal to distance divided by time. Substituting in our values, we have 1.5 is equal to š  sub two divided by 9.5. We can then multiply both sides of the equation by 9.5. And š  sub two is therefore equal to 14.25. In the 9.5 seconds, the cyclist traveled a distance of 14.25 meters. We can now calculate the total distance by adding 2.25 and 14.25. This is equal to 16.5. The total distance š  that the cyclist covered is 16.5 meters.

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