Question

question 4.
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?
a. ○ trial 1 speed = 2 m/s; trial 2 speed = 3 m/s → greater force increases 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: Calculate speed for Trial 1

Speed formula is \(v=\frac{d}{t}\). For Trial 1, \(d = 12\) meters and \(t=6\) seconds. So \(v_1=\frac{12}{6}=2\) m/s.

Step2: Calculate speed for Trial 2

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

Step3: Analyze the relationship between force and speed

Since speed \(v_2>v_1\) (3 m/s > 2 m/s) when the force (implied by greater distance in same time) is increased, greater force increases speed and distance.

Answer:

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