# Determine the time required for a car to travel

Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m/$s^2$ and decelerate at 2 m/$s^2$.

#### Solution:

Let us first represent the values we know in a diagram like so:

The total distance the car travels is a 1000 m. We also know that the car starts from rest, thus the initial velocity is 0 m/s and accelerates at 1.5 m/$s^2$. The maximum speed of the car is $v_{max}$. The car then decelerates at a rate of 2 m/$s^2$ and the velocity at the end is zero as the car comes to a stop.

let us use the following kinematic equation. We will be focusing only on the accelerating portion for now.

$v=v_0+at$

(Where $v$ is final velocity, $v_0$ is initial velocity, $a$ is acceleration and $t$ is time)

Let us substitute the values we know.

$v_{max}=0+(1.5)t_1$

$v_{max}=1.5t_1$ (eq.1)

Let us now use another kinematic equation.

$s=s_0+v_0t_1+\dfrac{1}{2}(a)(t_1)^2$

(Where $s$ is final displacement, $s_0$ is initial displacement, $v_0$ is initial velocity, $t$ is time, and $a$ is constant acceleration)

Substitute the values we know:

$x=0+0+\dfrac{1}{2}(1.5)(t_1)^2$

$x=\dfrac{1}{2}(1.5)(t_1)^2$ (eq.2)

Now, let us focus on the deceleration portion. As before, let us write our kinematics equations.

$0=v_{max}+(-2)t_2$ (eq.3)

(Remember, the initial velocity is now $v_{max}$ and the final velocity is 0 m/s, and our acceleration is negative because the car is slowing down)

Write the second kinematics equation as before:

$s=s_0+v_0t_1+\dfrac{1}{2}(a)(t_1)^2$

$1000-x=0+(v_{max})(t_2)+\dfrac{1}{2}(-2)(t_2)^2$

$1000-x=(v_{max})(t_2)-(t_2)^2$ (eq.4)

We can now simplify the equations by substituting the initial equations into the secondary equations we found. The goal here is to write $t_1$, $v_{max}$ and $x$ in terms of $t_2$.

Substitute eq.1 into eq.3:

$0=(1.5t_1)+(-2)t_2$

(Simplify)

$t_1=1.33t_2$

If $t_1=1.33t_2$ and $v_{max}=1.5t_1$ (eq.1), then $v_{max}=2t_2$

If $x=\dfrac{1}{2}1.5(t_1)^2$ (eq.2), and $t_1=1.33t_2$, then $x=1.33(t_2)^2$

Now that we have simplified our terms, we can substitute these values into eq.4, which yields:

$1000-1.33(t_2)^2=2(t_2)(t_2)-(t_2)^2$

$1000=2.33(t_2)^2$

$t_2=\sqrt{\dfrac{1000}{2.33}}$

$t_2=20.7$ s

Remember that $t_1=1.33t_2$, thus, $t_1=(1.33)(20.7)=27.5$ s.

Therefore, the total time is:

$t_{total}=20.7+27.5=48.2$ s