Question

A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 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 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

Answer 1

a) Since the speed is $V = a\times t$ , then $t_1=\frac{V}{a_1}$ , where $a_1$ is the acceleration in the first interval, $t_1$ is the time spent on the first interval, then

$t_1=\frac {20} 2=10$ s.

By condition $t_2=20$ c, and $t_3=5$ c, therefore, the total time is $t = 10 + 20 + 5 = 35$ s.

b) Average speed is determined by formula $V_{ave}=\frac {S_{total}}{t_{total}}$ ;

The path for the first interval is $S_1=\frac {a_1\times t_1^2}{2}=\frac {2\times 10 ^ 2} {2} = 100$ m;

For the second interval, the path is equal to $S_2=V\times t_2= 20\times 20 = 400$ m;

And for the third $S_3=\frac {V_1^2-V_0^2}{2\times a_2}$ , where $V_1=0$ , since this is the final speed at the stop, $V_0$ is the initial speed at the third interval, $a_2$ is the acceleration at the third interval and it is equal to $a_2=\frac {V_0}{t_3}=\frac {20} {5} = 4$ m/s $^ 2$ , since the acceleration is directed against the velocity vector then $a_2=-4$ m/s $^ 2$

$S_3=\frac {0-20^2}{-4\times 2}=50$ m;

As a result, we have:

$V_{ave}=\frac {400+100+50}{35}\approx15.71$ m/s.