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.


Show me the final answer↓

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

Determine the time required for a car to travel

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.


(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}=1.5t_1 (eq.1)


Let us now use another kinematic equation.


(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=\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:



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:




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:




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


Leave a comment

Your email address will not be published. Required fields are marked *