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 𝑠
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 𝑠 that he travels. We will split this into two parts:
𝑠 sub one, the journey on the hill, and 𝑠 sub two, the journey on the flat.
We can calculate the value of 𝑠
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 𝑠 sub one. We will use the equation 𝑣 squared
is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have
1.5 squared is equal to zero squared plus two multiplied by 0.5 multiplied by 𝑠 sub
one. The left-hand side is equal to
2.25. And as two multiplied by 0.5 is
one, this is equal to 𝑠 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 𝑠 sub two divided by 9.5. We can then multiply both sides of
the equation by 9.5. And 𝑠 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 𝑠 that the
cyclist covered is 16.5 meters.
