Answer to Question #129903 in Physics for Jack man

Question

A car travels 1km it starts to accelerate at 2.5m/s it attains a velocity of 12.5m/s the car continues at this velocity for some time and decelerates at 3m/s^2 until it stops. What is the total time for the whole distance?

Answer 1

Let the first acceleration be $a_1 = 2.5 m/s^2$ and the second one $a_2 = 3m/s^2$. The Initial speed is $v_0 = 0$, the next one is $v_1 = 12.5m/s$ and the final is $v_2 = 0$. The total distance is $d = 1km = 1000m$.

Let's find the time $t_1$ of the first acceleration from $v_0$ to $v_1$

By definition, the acceleration is:

$a_1 = \dfrac{v_1 - v_0}{t_1}\\ t_1 = \dfrac{v_1 - v_0}{a_1} = \dfrac{12.5-0}{2.5} = 5s$

During this time the car covered the distance:

$d_1 = \dfrac{a_1t_1^2}{2} = \dfrac{2.5\cdot 5^2}{2} = 31.25 m$

Let's find the time $t_3$ of the last acceleration. Again, by definition:

$a_2 = \dfrac{v_2-v_1}{t_3}\\ t_1 = \dfrac{|v_2 - v_1|}{a_2} = \dfrac{|0-12.5|}{3} = \dfrac{25}{6}s$

During this time the car covered the distance:

$d_3 = \dfrac{a_2t_3^2}{2} = \dfrac{3\cdot 25^2}{2\cdot 6^2} = 156.25 m$

The distance in the middle, covered with a constant speed $v_1$ is then:

$d_2 = d-d_1 - d_3 = 1000m - 31.25m - 156.25m = 812.5m$

Thus, the time during which it was covered is:

$t_2 = d_2/v_1 = 65 s$

Thus, the total time is:

$t = t_1 + t_2 + t_3 = 5s + \dfrac{25}{6}s + 65s \approx 74.17s$

Answer. 74.17s.