Question

motion quiz a
show all work!

1. (20 points) a car travels east with a speed of 35.0 m/s for 1.52 minutes.

how far does this car move and in what direction?
85.0 meters west in 17.55 seconds, turns around and runs
53.0 meters in 10.14 seconds.

Explanation:

Step1: Convert time to seconds

\(t_1 = 1.52\times60\ s = 91.2\ s\)

Step2: Calculate distance traveled east

\(d_1=v_1t_1=35.0\times91.2\ m = 3192\ m\)

Step3: Calculate total distance traveled west

\(d_{west}=d_2 + d_3=85.0 + 53.0\ m=138\ m\)

Step4: Calculate net displacement

\(d_{net}=d_1 - d_{west}=3192-138\ m = 3054\ m\) (east)

Step5: Calculate total distance

\(D=d_1 + d_2 + d_3=3192+85.0+53.0\ m = 3330\ m\)

Answer:

1. First, convert the time of the first - part motion to seconds. Since 1 minute = 60 seconds, for the first part of the motion where the car travels east with a speed \(v_1 = 35.0\ m/s\) for \(t_1=1.52\) minutes, \(t_1 = 1.52\times60\ s=91.2\ s\).

• Using the formula \(d = vt\) (where \(d\) is distance, \(v\) is speed, and \(t\) is time), the distance \(d_1\) traveled east is \(d_1=v_1t_1\).
• Substitute \(v_1 = 35.0\ m/s\) and \(t_1 = 91.2\ s\) into the formula: \(d_1=35.0\times91.2\ m = 3192\ m\).
• The car then travels west a distance \(d_2 = 85.0\ m\) in \(t_2 = 17.55\ s\), and then travels a distance \(d_3 = 53.0\ m\) in \(t_3 = 10.14\ s\).
• The net displacement in the east - west direction:
• The total distance traveled east is \(d_1\), and the total distance traveled west is \(d_{west}=d_2 + d_3=85.0+53.0\ m = 138\ m\).
• The net displacement \(d_{net}=d_1 - d_{west}=3192-138\ m = 3054\ m\) to the east.
• The total distance traveled (the path - length) \(D=d_1 + d_2 + d_3=3192+85.0 + 53.0\ m=3330\ m\).

1. The direction of the net displacement is east.

The distance the car moves (total path - length) is \(3330\ m\) and the net displacement is \(3054\ m\) in the east direction.