A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east
ward component of the car’s velocity is given by vx(t)=(0.860m/s3)t2. What is the acceleration of the car when vx=12.0m/s?
1 answer:
Answer:
ax = 6.43m/s²
Explanation:
The acceleration is the time derivative of the velocity function ax = dvx(t)/dt
We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.
So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.
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Answer: 2.04 s
Explanation:
Let the initial velocity be v, Angle of projectile be
Then the horizontal component = v cos θ = 16 m/s
Vertical component of velocity = v sin θ = 20 m/s
Time taken to reach the highest point is half the time taken for total flight.
Time for total flight,
Thus, the football takes 2.04 s to rise to the highest point of its trajectory.