A cyclist is coasting up a hill. At t=0, the cyclist has a velocity of 14.27 m/s. If the acceleration of
the cyclist is constant at -1.67 m/s2, how long will it take the cyclist to come to a complete stop? If the
peak of the hill is 40 m away from the cyclist at t=0, will she have to pedal to reach the top?
how long does it take for v to drop to zero?
v(t) = 14.27 - 1.67t
using that t, the distance moved is
s(t) = 14.27t - 0.835t^2
To find out how long it will take for the cyclist to come to a complete stop, we can use the equation:
vf = vi + at
Where:
vf = final velocity (which is 0 m/s since the cyclist comes to a complete stop)
vi = initial velocity (given as 14.27 m/s)
a = acceleration (given as -1.67 m/s^2)
t = time
Rearranging the equation to solve for time (t), we have:
t = (vf - vi) / a
Substituting the given values, we get:
t = (0 - 14.27) / (-1.67)
t = 8.54 seconds (rounded to two decimal places)
Therefore, it will take the cyclist approximately 8.54 seconds to come to a complete stop.
To determine if the cyclist will have to pedal to reach the top of the hill, we need to calculate the distance the cyclist will travel during the time it takes to come to a complete stop.
Using the equation of motion, we have:
d = vi * t + 0.5 * a * t^2
Where:
vi = initial velocity (given as 14.27 m/s)
t = time (8.54 seconds, as calculated above)
a = acceleration (given as -1.67 m/s^2)
d = distance traveled
Substituting the values, we get:
d = 14.27 * 8.54 + 0.5 * (-1.67) * (8.54)^2
d = 122.1 meters (rounded to one decimal place)
Since the distance to the peak of the hill is 40 meters, the cyclist will not have to pedal to reach the top.
To find the time it will take for the cyclist to come to a complete stop, we can use the kinematic equation:
vf = vi + at
where:
vf = final velocity (which is 0 m/s, since the cyclist comes to a complete stop)
vi = initial velocity (14.27 m/s)
a = acceleration (-1.67 m/s^2)
t = time
Plugging in the values we have, we get:
0 = 14.27 + (-1.67)t
To solve for t, we can rearrange the equation:
1.67t = 14.27
t = 14.27 / 1.67
t ≈ 8.56 seconds
So, it will take approximately 8.56 seconds for the cyclist to come to a complete stop.
To determine whether the cyclist will have to pedal to reach the top of the hill, we need to calculate the distance covered during the deceleration period (from t=0 to when the cyclist comes to a stop). We can use the kinematic equation:
d = vit + (1/2)at^2
where:
d = distance covered
vi = initial velocity (14.27 m/s)
a = acceleration (-1.67 m/s^2)
t = time taken to come to a stop (8.56 seconds)
Plugging in the values we have, we get:
d = (14.27)(8.56) + (1/2)(-1.67)(8.56)^2
d ≈ 122.1912 meters
Since the peak of the hill is 40 meters away from the cyclist at t=0, and the distance covered during deceleration is greater than the distance to the peak of the hill, the cyclist will not have to pedal to reach the top.