A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.

4592.57 lb

Explanation:

The missing diagram for this question is attached in the image below.

Given that:

the weight of the car = 2890 lb

At point A, the speed of the car = 58 mi/hr

At point C, the speed of the car = 18 mi/hr

To ft/s:

 = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

 = 85.07 ft/s

= 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

= 26.4 ft/s

Between A to C, the total distance is;

Now, we need to determine the deceleration of the car using the formula:

a = -4.99 ft/s²

The velocity of the car as it passes via B

Along B, the car's acceleration is:

Finally, the total horizontal force F exerted = m

= 4592.57 lb

compass can be used day and night, cloudy or sunny

force tends to be low and long diustance masses are high