Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to t
1 answer:
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Answer:
a)
Explanation:
- Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:
- Since the car starts from rest, v₀ =0.
- We know the value of t = 5 sec., but we need to find the value of a.
- Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:
- Replacing a and t in (1):
- Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
- Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:
- Replacing v₀, at and t in (1), we have:
- Therefore, as the truck travels twice as far as the car, the right answer is a).