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Tasya [4]
3 years ago
13

Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t

he car? (b) twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?
Physics
1 answer:
Stels [109]3 years ago
6 0

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

a =\frac{ v-u}{\Delta t}

Here, v is final velocity and u is the initial velocity and \Delta t is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2

Thus, the average acceleration of the car is 14 .3 \ m/s^2 in the direction to the left.

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1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
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1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

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F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

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However, in this case the displacement is

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Therefore, the magnitude of the force in this case is

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