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:
Velocity of truck is 21.875 m/s.
Explanation:
Given:
Mass of the car (m) = 2500 kg
Initial speed of the car (u) = 35 m/s
Mass of the truck (M) = 4000 kg
Initial speed of the truck (U) = 0 m/s (Rest)
As per question, the total momentum of the car gets transferred to the truck.
Therefore, the final momentum of the car is 0.
Momentum is given as the product of mass and velocity.
So, if momentum becomes zero means the velocity becomes 0.
Therefore, final velocity of the car (v) = 0 m/s
Let the final velocity of truck be 'V' m/s.
Now, during collision, the total momentum remains conserved.
Initial momentum = Final momentum.
Initial momentum of car + Initial momentum of truck = Final momentum of car + Final momentum of truck.
⇒
Therefore, the velocity of the truck is 21.875 m/s.