A bicyclist accelerates -0.87 m/s/s during a 3.8 s interval. What is the change in velocity of the bicyclist and the bicycle?
For this problem, I used the formula 'change in v = a x t" and came up with -3.306 m/s as my answer. Is this right?
correct.
Yes, your answer is correct. The formula for calculating the change in velocity is:
change in v = a x t
where:
a = acceleration
t = time interval
In this case, the acceleration is given as -0.87 m/s^2 and the time interval is 3.8 s. Substituting these values into the formula, we get:
change in v = (-0.87 m/s^2) x (3.8 s) = -3.306 m/s
So, the change in velocity of the bicyclist and the bicycle is -3.306 m/s. The negative sign indicates that the direction of the velocity is decreasing.
Yes, your answer is correct. To find the change in velocity of the bicyclist and the bicycle, you used the correct formula, which is change in velocity (Δv) equals acceleration (a) multiplied by time (t).
Given:
Acceleration (a) = -0.87 m/s²
Time interval (t) = 3.8 s
Using the formula Δv = a x t, you can substitute the given values:
Δv = (-0.87 m/s²) × (3.8 s)
Now, multiplying these values gives:
Δv = -3.306 m/s
Therefore, the change in velocity of the bicyclist and the bicycle is -3.306 m/s. The negative sign indicates that the velocity is decreasing, which means that the bicyclist is slowing down.