True.
Depending how accurate the graph is plotted
Answer:
t = 3 seconds
Explanation:
Given that,
Initial speed, u = 30 m/s
Final speed, v = 0
It slides 45 m to rest.it take the box to come to rest
We need to find how long it take the box to come to rest.
Let a be the acceleration and t is time.

Now finding time.

So, the required time is 3 seconds.
That would be a frequency of 1.2666... beats per second. This can be phrased as your heart beats at 1.27 Hz.
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:

- radius of the hill:

Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,

, so we can write:

(1)
By rearranging the equation and substituting the numbers, we find N:

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:

from which we find