Answer to Question #188524 in Physics for Stacey Chua

Question

What is the final velocity of a 1200 kilogram car starting from rest after 50 meters when 4500 Newton's of force is applied?

Answer 1

According to the second Newton's law, the acceleration of the car is:

$$a = \dfrac{F}{m}$$

where $F = 4500N$ is the total force applied to the car and $m = 1200kg$ is the mass of the car.

According to the kinematic equation, the distance covered under a constant acceleration is:

$$d =v_0t + \dfrac{at^2}{2} = \dfrac{at^2}{2}$$

where $v_0 = 0$ is the initial speed of the car (equal to 0 since the car starts from rest) and $t$ is the time.

Since $d = 50m$, we can find the time:

$$t = \sqrt{\dfrac{2d}{a}} = \sqrt{\dfrac{2dm}{F}} \\$$

The velocity of the car at this moment of time (if it starts from rest) is given by the following kinematic equation:

$$v =v_0 + at = at$$

Substituting the expressions for $a$ and $t$, obtain:

$$v = \dfrac{F}{m}\cdot \sqrt{\dfrac{2dm}{F}} = \sqrt{\dfrac{2dF}{m}}\\ v = \sqrt{\dfrac{2\cdot 50\cdot 4500}{1200}} \approx 6.1 m/s$$

Answer. 6.1 m/s.