Question

) A truck on a straight road starts from rest and accelerates at 1.0 m/s2 until it reaches a speed of 10 m/s. Then the truck travels for 30 s at the constant speed of 10 m/s 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?

Answer 1

Answer on Question #40184 – Physics – Mechanics

1. A truck on a straight road starts from rest and accelerates at $1.0 \, \text{m/s}^2$ until it reaches a speed of $10 \, \text{m/s}$. Then the truck travels for $30 \, \text{s}$ at the constant speed of $10 \, \text{m/s}$ until the brakes are applied, stopping the car in a uniform manner in an additional $5.0 \, \text{s}$. How long is the truck in motion and what is its average velocity during the motion?

The total time of the movement: $t = t_1 + t_2 + t_3$, $\boxed{t = \frac{v}{a_1} + t_2 + t_3}$.

The average velocity during the motion of the truck: $\overline{v} = \frac{l_1 + l_2 + l_3}{t_1 + t_2 + t_3}$, $\overline{v} = \frac{\frac{v^2}{2a_1} + vt_2 + \frac{vt_3}{2}}{\frac{v}{a_1} + t_2 + t_3}$.

Let check the dimensions.

$$[t] = \frac{m/s}{m/s^2} + s = s, \quad [\overline{v}] = \frac{\frac{(m/s)^2}{m/s^2} + \frac{m}{s} \cdot s}{\frac{m/s}{m/s^2} + s} = \frac{m}{s}$$

Let evaluate the quantities.

$$t = \frac{10}{1} + 30 + 5 = 45(s), \quad \overline{v} = \frac{\frac{10^2}{2 \cdot 1} + 10 \cdot 30 + \frac{10 \cdot 5}{2}}{\frac{10}{1} + 30 + 5} \approx 8.3\left(\frac{m}{s}\right).$$

Answer: $45s$, $8.3\frac{m}{s}$.