Question

a truck on a straight road start from rest accelerating at 2 m/s^2 until it reaches a speed of 20 m/s then the truck travels for 20 s at a constant speed until the brakes are applied stopping the truck in a uniform manner in an additional 5 s . how long is the truck in motion and what is the average velocity of the truck for the motion described ?

Answer 1

A truck on a straight road starts from rest and accelerates at $2.0 \, \text{m/s}^2$ until it reaches a speed of $20 \, \text{m/s}$. Then the truck travels for 20 s at a constant speed until the brakes are applied, stopping the car in a uniform manner in an additional 5.0 s. How long is the truck in motion and what is its average velocity during the motion?

Solution

Assumption: to find the average speed we can break all path up into three phases: accelerating $\rightarrow$ coasting $\rightarrow$ decelerating.

First phase (accelerating)

Rate equation for the acceleration phase:

$$V_1 = a_1 t_1$$ $$t_1 = \frac{V_1}{a_1} = \frac{20 \frac{\text{m}}{\text{s}}}{2.0 \frac{\text{m}}{\text{s}^2}} = 10 \, \text{s}$$

It means that accelerating will stop after 10 seconds because then the truck will reach velocity $20 \, \text{m/s}$.

Equation of motion for the acceleration phase:

$$S_1 = \frac{a_1 t_1^2}{2} = \frac{2.0 \frac{\text{m}}{\text{s}^2} \cdot (10 \, \text{s})^2}{2} = 100 \, \text{m}$$

Second phase (coasting):

Truck travels path at $20 \, \text{m/s}$ for 20 s:

$$S_2 = V_2 t_2 = 20 \frac{\text{m}}{\text{s}} \cdot 20 \, \text{s} = 400 \, \text{m}$$

Third phase (decelerating):

Rate equation for the acceleration phase:

$$0 = V_2 - a_3 t_3$$ $$a _ {3} = \frac {V _ {2}}{t _ {3}} = \frac {2 0 \frac {\mathrm {m}}{\mathrm {s}}}{5 \mathrm {s}} = 4 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}$$

It means that accelerating will stop after 10 seconds because then the truck will reach velocity $20\mathrm{m / s}$.

Equation of motion for the deceleration phase:

$$S _ {3} = \frac {a _ {3} t _ {3} ^ {2}}{2} = \frac {4 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot (5 \mathrm {s}) ^ {2}}{2} = 5 0 \mathrm {m}$$ $$\begin{array}{l} \text{Average velocity} = \frac {\text {all distance}}{\text {all time}} = \frac {S _ {1} + S _ {2} + S _ {3}}{t _ {1} + t _ {2} + t _ {3}} = \frac {1 0 0 m + 4 0 0 m + 5 0 m}{1 0 s + 2 0 s + 5 s} = \\ = 1 5. 7 \frac {\mathrm {m}}{\mathrm {s}} \\ \end{array}$$

Time of the travel is $T = 10s + 20s + 5s = 35s$.

Answer: average velocity during the motion is $15.7\frac{\mathrm{m}}{\mathrm{s}}$, truck was in motion during 35s.