With an average acceleration of negative 1.2m/s squared, how long will it take a cyclist to bring a bicylcle with an initial speed of 6.5 m/s to a complete stop?
answers are:
either 2.0, 5.4. 0.18 or -1.8s
Please help-I am totally confused
To find the time it takes for the cyclist to bring the bicycle to a complete stop, we can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s, as the bicycle comes to a complete stop)
u = initial velocity (6.5 m/s)
a = acceleration (-1.2 m/s^2)
t = time
Rearranging the equation to solve for time (t), we get:
t = (v - u) / a
Substituting the given values:
t = (0 - 6.5) / (-1.2)
t = (-6.5) / (-1.2)
t ≈ 5.42 seconds
Therefore, the time it will take for the cyclist to bring the bicycle to a complete stop is approximately 5.42 seconds.
Based on the options provided, the closest answer is 5.4s.
To solve this problem, we can use the kinematic equation:
v^2 = u^2 + 2aΔs
Where:
- v is the final velocity (0 m/s, since the cyclist comes to a complete stop)
- u is the initial velocity (6.5 m/s)
- a is the average acceleration (-1.2 m/s^2)
- Δs is the displacement (which we don't need to find)
Plug in the given values into the equation:
0^2 = (6.5)^2 + 2(-1.2)Δs
Simplifying the equation further:
0 = 42.25 - 2.4Δs
Rearranging the equation to solve for Δs:
2.4Δs = 42.25
Δs = 42.25 / 2.4
Δs ≈ 17.6 meters
Now, we can use the equation of motion:
v = u + at
Since v = 0 m/s (final velocity), u = 6.5 m/s (initial velocity), and a = -1.2 m/s^2 (average acceleration), we can solve for t:
0 = 6.5 + (-1.2)t
-6.5 = -1.2t
Dividing both sides by -1.2:
t ≈ 5.4 seconds
Therefore, the correct answer is 5.4 seconds.