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:

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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

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

Solution:

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