I NEED HELP PLEASE. :'((((

Question

The table below shows the elapsed time when two different cars pass a 10, 20, 30, 40 and 50-meter mark on a test track. 

a) For car 1, what is the average velocity (change in distance divided by change in time) between the 0 and 10-meter mark? Between the 0 and 50-meter mark? Between the 20 and 30-meter mark? Analyze the data to describe the motion of car 1. 

b) How does the velocity of car 1 compare to that of car 2?

	CAR 1	CAR 2
d	t	t
10	4.472	1.742
20	6.325	2.899
30	7.746	3.831
40	8.944	4.633
50	10.000	5.348

 

Answer 1

a) CAR 1 av speed = 10/4.472=2.24 m/s approx (0-10m); 50/10=5 m/s (0-50m); (30-20)/(7.746-6.325)=10/1.421=7.04 m/s. The average speed is increasing so the car is accelerating. 

b) CAR 2 av speed=10/1.742=5.74 (0-10m); 50/5.348=9.35 m/s (0-50m); 10/(0.932)=10.73 m/s. CAR 2 has greater average speeds than CAR 1. Both cars are accelerating. 

The accelerations can be measured by calculating the change in speed over time. A graph or two would be useful in comparing the performance of the two cars.