Question

while shopping with her parents, mona pushes a cart across the stores smooth floor. she records the distance the cart travels and the time it takes.
trial 1: the cart travels 12 meters in 6 seconds.
trial 2: the cart travels 18 meters in 6 seconds.
what is the carts speed in each trial, and what does this suggest about the effect of monas applied force on constant - speed motion?
a. trial 1 speed = 2 m/s; trial 2 speed = 3 m/s - greater force increased speed and distance.
b. trial 1 speed = 2 m/s; trial 2 speed = 3 m/s - speed stays the same no matter the applied force.
c. trial 1 speed = 6 m/s; trial 2 speed = 12 m/s - doubling force always doubles speed.
d. trial 1 speed = 12 m/s; trial 2 speed = 18 m/s - distance is the only factor that matters.

Explanation:

Step1: Recall speed formula

Speed \(v=\frac{d}{t}\), where \(d\) is distance and \(t\) is time.

Step2: Calculate speed for Trial 1

For Trial 1, \(d_1 = 12\) meters and \(t_1=6\) seconds. So \(v_1=\frac{d_1}{t_1}=\frac{12}{6}=2\) m/s.

Step3: Calculate speed for Trial 2

For Trial 2, \(d_2 = 18\) meters and \(t_2 = 6\) seconds. So \(v_2=\frac{d_2}{t_2}=\frac{18}{6}=3\) m/s.

Answer:

A. Trial 1 speed = 2 m/s; Trial 2 speed = 3 m/s - Greater force increased speed and distance.