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:

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.

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

 

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