Question

retake velocity assessment
if the distance is in kilometers and time is in hours;
the velocity of the car from t = 4.30 to t = 4.45 hours is
the velocity of the car from t = 5.00 to t = 5.15 hours is
the velocity of the car from t = 5.15 to t = 6.15 hours is

Explanation:

Step1: Recall velocity formula

Velocity $v=\frac{\Delta d}{\Delta t}$, where $\Delta d$ is change in distance and $\Delta t$ is change in time.

Step2: Calculate for $t = 4.30$ to $t = 4.45$

From the graph, at $t = 4.30$ distance $d_1 = 30$ km, at $t = 4.45$ distance $d_2=30$ km. $\Delta t=4.45 - 4.30=0.15$ hours, $\Delta d=d_2 - d_1=30 - 30 = 0$ km. So $v=\frac{0}{0.15}=0$ km/h.

Step3: Calculate for $t = 5.00$ to $t = 5.15$

At $t = 5.00$ distance $d_1 = 50$ km, at $t = 5.15$ distance $d_2 = 50$ km. $\Delta t=5.15 - 5.00 = 0.15$ hours, $\Delta d=d_2 - d_1=0$ km. So $v=\frac{0}{0.15}=0$ km/h.

Step4: Calculate for $t = 5.15$ to $t = 6.15$

At $t = 5.15$ distance $d_1 = 50$ km, at $t = 6.15$ distance $d_2 = 0$ km. $\Delta t=6.15 - 5.15 = 1$ hour, $\Delta d=0 - 50=- 50$ km. So $v=\frac{-50}{1}=-50$ km/h.

Answer:

The velocity of the car from $t = 4.30$ to $t = 4.45$ hours is $0$ km/h.
The velocity of the car from $t = 5.00$ to $t = 5.15$ hours is $0$ km/h.
The velocity of the car from $t = 5.15$ to $t = 6.15$ hours is $- 50$ km/h.