Question

a dog named rascal leaves his home and travels a short distance and then returns home. the graph shows his position as a function of time. how does his velocity on his return trip compare to his velocity for the first 5 seconds? a his velocity was slower when compared to the first 5 seconds. b his velocity was positive on the return trip when compared to the first 5 seconds. c his velocity on the return trip was twice as slow when compared to the first 5 seconds. d his velocity on the return trip was twice as fast when compared to the first 5 seconds.

Explanation:

Step1: Recall velocity - position relationship

Velocity is the slope of the position - time graph.

Step2: Calculate velocity for first 5 seconds

For the first 5 seconds, the position changes from 0 m to 5 m in 5 s. Using the slope formula $v_1=\frac{\Delta x}{\Delta t}$, we have $v_1=\frac{5 - 0}{5}=1$ m/s.

Step3: Calculate velocity for return trip

The return trip occurs from $t = 10$ s to $t = 13$ s, and the position changes from 5 m to 0 m. Using the slope formula $v_2=\frac{\Delta x}{\Delta t}$, we have $v_2=\frac{0 - 5}{13 - 10}=-\frac{5}{3}\approx - 1.67$ m/s. The magnitude of $v_2$ is $\frac{5}{3}$ m/s and the magnitude of $v_1$ is 1 m/s. The magnitude of $v_2$ is approximately 1.67 times the magnitude of $v_1$, which means the velocity on the return trip is faster. Also, the velocity on the first 5 - second part is positive (moving away from home) and on the return trip is negative (moving towards home).

Answer:

D. His velocity on the return trip was twice as fast when compared to the first 5 seconds.