Question

What is the final velocity of a car that is originally traveling 12 m/s and then undergoes an acceleration of 2.3m/s squared for a distance of 130m?

Answer 1

To solve this problem we use the general kinetic equations.

We need to know the time it takes for the car to reach 130 meters.

In this way we have to:

`x(t) = x_0 + v_0t + 0.5at ^ 2`

Where

`x_0` = initial position

`v_0` = initial velocity

`a` = acceleration

`t` = time

`x(t)` = position as a function of time

`130 = 0 + 12(t) + 0.5(2.3)t ^ 2`

`1.15t ^ 2 + 12t - 130`.

We use the quadratic formula to solve the equation.

`t = \frac{-12 \± \sqrt {(12) ^ 2-4(1.15)(- 130)}}{2 (1.15)}`

t = 6.63 s and t = -17.1 s

We take the positive solution. This means that the car takes 6.63 s to reach 130 meters.

Then we use the following equation to find the final velocity:

`v_f = v_0 + at`

Where:

`v_f` = final speed

`v_f = 12 +2.23(6.63)`

The final speed of the car is 27.25 m/s