With an average acceleration of −0.96 m/s2,
how long will it take a cyclist to bring a bicycle
with an initial speed of 14.8 m/s to a complete
stop?
Answer in units of s
To find out how long it will take for the cyclist to bring the bicycle to a complete stop, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
- v is the final velocity (0 m/s, since the bicycle comes to a complete stop)
- u is the initial velocity (14.8 m/s)
- a is the acceleration (-0.96 m/s^2, note the negative sign indicating deceleration)
- s is the displacement (we're trying to find this)
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the given values, we have:
s = (0^2 - 14.8^2) / (2 x -0.96)
Now we can solve for s:
s = (-219.04) / (-1.92)
s ≈ 113.95 m
The displacement s represents the distance the bicycle will travel until it comes to a complete stop. However, the question asks for the time it takes to stop, so we need to find the time. We can use another kinematic equation:
v = u + at
where:
- v is the final velocity (0 m/s)
- u is the initial velocity (14.8 m/s)
- a is the acceleration (-0.96 m/s^2)
- t is the time (we're trying to find this)
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (0 - 14.8) / (-0.96)
Now we can solve for t:
t = (-14.8) / (-0.96)
t ≈ 15.42 s
Therefore, it will take approximately 15.42 seconds for the cyclist to bring the bicycle with an initial speed of 14.8 m/s to a complete stop.