Consider the car driving around a circular track. Once started, the driver maintains a speed of 45 m/h. Compare the
driver's velocity with his acceleration in this case.
The car is accelerating because it is changing direction but the velocity is
constant. the
A)
Eliminate
The car is traveling at a constant rate of speed so it is not accelerating and
the velocity does not change.
Acceleration is the change in velocity so the car is experiencing a change
in velocity and it is accelerating.
The car is traveling at a constant rate of speed so it is not accelerating but
the velocity changes because the direction the car travels is constantly
changing.
I

Acceleration is the change in velocity so the car is experiencing a change

in velocity and it is accelerating.

Explanation:

- Velocity is the rate of change of the position of an object per unit time. It is a vector quantity, so it has both a magnitude (the speed) and a direction:

where is the change in position while is the time elapsed.

- Acceleration is the rate of change of velocity per unit time. Similarly, acceleration is also a vector, so it has both a magnitude and a direction:

where is the change in velocity while is the time elapsed.

Since velocity is a vector, (change in velocity) is also a vector. This means that there can be a change in velocity if either two of the conditions occur:

- A change in the speed

- A change in the direction of the velocity

For an object in uniform circular motion, the speed remains constant, however the velocity keeps changing, since the direction of the velocity keeps changing. This means that a car in uniform circular motion is experiencing a non-zero acceleration.

So, the correct option is

Acceleration is the change in velocity so the car is experiencing a change

in velocity and it is accelerating.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j