Question

5. Law of Acceleration: A 1400 kg car is initially at rest. It is then accelerated by a 3000 N force for 10 seconds.What is the car's acceleration?How fast is it going after 10 seconds?

Answer 1

The force acting on the car can be expressed as,

$F=ma$

Plug in the known values,

$\begin{gathered} 3000\text{ N=(1400 kg)a} \\ a=\frac{3000\text{ N}}{1400\text{ kg}}(\frac{1kgm/s^2}{1\text{ N}}) \\ =2.14m/s^2 \end{gathered}$

Thus, the acceleration of the car is 2.14 m/s2.

The final velocity of the car is given as,

$v=u+at$

Substitute the known values,

$\begin{gathered} v=0m/s+(2.14m/s^2)(10\text{ s)} \\ =21.4\text{ m/s} \end{gathered}$

Thus, the final velocity of the car is 21.4 m/s.