With an average acceleration of –0.5 m/s2, (a) how long will it take a cyclist to bring a bicycle with an initial velocity of +13.5 m/s to a complete stop? (b) Calculate the distance required for the bicyclist to come to rest. (c) If the cyclist’s initial speed is doubled to 27 m/s under the same condition, calculate the distance required for the cyclist to come to a halt.
(a) Well, if the cyclist starts with a velocity of +13.5 m/s and has an average acceleration of -0.5 m/s^2, we can use the equation v = u + at to find the time it takes to stop. Since the final velocity v will be 0 m/s, we can rearrange the equation to solve for time t: t = (v - u) / a.
Plugging in the values, we get t = (0 - 13.5) / -0.5 = 27 s. So, it will take the cyclist 27 seconds to come to a complete stop.
(b) To calculate the distance required for the bicyclist to come to rest, we can use another equation of motion: s = ut + (1/2)at^2. Since the final velocity is 0, we can ignore the (1/2)at^2 term, and we're left with s = ut.
Plugging in the values, we have s = 13.5 m/s * 27 s = 364.5 m. So, the distance required for the bicyclist to come to rest is 364.5 meters.
(c) Now, let's see what happens when the cyclist's initial speed is doubled to 27 m/s. Using the same average acceleration of -0.5 m/s^2, we can use the equation s = ut + (1/2)at^2.
Since the final velocity is 0 m/s, we have 0 = 27 m/s * t + (1/2) * -0.5 m/s^2 * t^2. Simplifying the equation, we get t^2 - 54t = 0.
Solving for t, we find t = 0 or t = 54 s. Since time can't be negative, we discard t = 0. Thus, it will take the cyclist 54 seconds to stop.
To find the distance required, we can use the equation s = ut + (1/2)at^2.
Plugging in the values, we have s = 27 m/s * 54 s + (1/2) * -0.5 m/s^2 * (54 s)^2 = 729 m. So, the distance required for the cyclist to come to a halt when starting at 27 m/s is 729 meters.
Hope that helps, and remember, keep biking and stay safe!
(a) To find the time it takes for the cyclist to bring the bicycle to a complete stop, we can use the equation:
v = u + at
where:
v = final velocity (0 m/s, since the cyclist comes to a complete stop)
u = initial velocity (13.5 m/s)
a = acceleration (-0.5 m/s^2)
t = time
Rearranging the equation to solve for time:
t = (v - u) / a
t = (0 - 13.5) / (-0.5)
t = 27 seconds
Therefore, it will take 27 seconds for the cyclist to bring the bicycle to a complete stop.
(b) To calculate the distance required for the bicyclist to come to rest, we can use the equation:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity (13.5 m/s)
t = time (27 seconds)
a = acceleration (-0.5 m/s^2)
Plugging in the values:
s = (13.5 * 27) + (0.5 * (-0.5) * (27)^2)
s = 364.5 - 182.25
s = 182.25 meters
Therefore, the distance required for the bicyclist to come to rest is 182.25 meters.
(c) If the cyclist's initial speed is doubled to 27 m/s under the same conditions, we can use the same equation as in part (b):
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity (27 m/s)
t = time
a = acceleration (-0.5 m/s^2)
Using the same acceleration as before (-0.5 m/s^2), we can find the time it takes to come to a halt:
0 = 27 + (-0.5)t
t = -27 / (-0.5)
t = 54 seconds
Now, we can calculate the distance:
s = (27 * 54) + (0.5 * (-0.5) * (54)^2)
s = 1458 - 729
s = 729 meters
Therefore, if the cyclist's initial speed is doubled to 27 m/s, the distance required for the cyclist to come to a halt is 729 meters.
To solve this problem, we need to use the equations of motion under constant acceleration. We'll start with part (a) of the question.
(a) To find the time it takes for the cyclist to bring the bicycle to a complete stop, we can use the equation:
v = u + at
where:
v = final velocity (0 m/s, as the bicycle comes to a complete stop)
u = initial velocity (13.5 m/s, given)
a = acceleration (-0.5 m/s^2, given)
t = time (unknown)
Rearranging the equation to solve for time, we get:
t = (v - u) / a
Substituting the given values, we have:
t = (0 - 13.5) / (-0.5)
t = 27 seconds
Therefore, it will take the cyclist 27 seconds to bring the bicycle to a complete stop.
(b) To calculate the distance required for the bicyclist to come to rest, we can use the equation:
s = ut + (1/2)at^2
where:
s = distance (unknown)
u = initial velocity (13.5 m/s, given)
a = acceleration (-0.5 m/s^2, given)
t = time (27 seconds, calculated in part (a))
Substituting the given values, we have:
s = (13.5 * 27) + (1/2) * (-0.5) * (27)^2
s = 364.5 - (0.5) * (0.5) * 729
s = 364.5 - 364.5
s = 0 meters
Therefore, the distance required for the bicyclist to come to rest is 0 meters.
(c) If the cyclist's initial speed is doubled to 27 m/s under the same condition, we need to calculate the distance required for the cyclist to come to a halt.
Using the same equation as in part (b), we have:
s = ut + (1/2)at^2
where:
s = distance (unknown)
u = initial velocity (27 m/s, given)
a = acceleration (-0.5 m/s^2, given)
t = time (unknown)
Rearranging the equation, we get:
s = (u^2 - v^2) / (-2a)
Substituting the given values, we have:
s = (27^2 - 0^2) / (-2 * -0.5)
s = 729 / 1
s = 729 meters
Therefore, if the cyclist's initial speed is doubled, it would require 729 meters for the cyclist to come to a halt.
(a) time = velocity / acceleration
(b) distance = average velocity * stopping time
... average velocity = initial velocity / 2 ... final velocity is zero
(c) doubling the speed will double the average velocity
... it will also double the stopping time
... so the stopping distance is quadrupled