For the instant represented, car A has an acceleration in the direction of its motion, and car B has a speed of 45 mi/hr which is increasing. If the acceleration of B as observed from A is zero for this instant, determine the magnitude of the acceleration of A and the rate at which the speed of B is changing.

Explanation:

Firstly, there is supposed to be a diagram attached in order  to complete this question;

I have attached  the diagram below in order to solve this question.

From the data given;

The radius of the car R = 600 ft

Velocity of the car  B,

We are to determine  the magnitude of the acceleration of A and the rate at which the speed of B is changing.

To start with the magnitude of acceleration A;

We all know that

1 mile = 5280 ft and an hour =  3600 seconds

Thus for ; 1 mile/hr ; we have :

5280 ft/ 3600 seconds

= 22/15 ft/sec

However;

for the velocity of the car B = 45 mi/hr; to ft/sec, we have:

= (45 × 22/15) ft/sec

= 66 ft/sec

A free body diagram is attached in the second diagram showing how we resolve the vector form

Now; to determine the magnitude of the acceleration of A; we have:

Where;

= radial acceleration of B

= tangential acceleration of B

From observation in the diagram; The acceleration of B is 0 from A

So;

Equating the coefficient of i and j now; we have :

From equation (2)

replace with 7.26 ft/s^2; we have

Similarly;

From equation (1)

replace with 10.267 ft/s^2

5.3m/; 7.45m/

Explanation:

the magnitude of a=(change in speed)/(the time took to accelerate to the speed above)

So for that sports car, a=(26.8m/s-0m/s)/(5.1s)=5.2549m/, knowing from the question that the sig fig of this question depend on "5.1 seconds", so the answer should be 5.3 m/

For the rocket, a=(447m/s-0m/s)/(1 minute)=(447m/s)/(60s)=7.45m/