This is a sample interview question from Oxford:

A ball, initially at rest, is pushed upwards by a constant force for a certain amount of time. Sketch the velocity of the ball as a function of time, from start to when it hits the ground.

My understanding is that the force is continuously applied since the force is "constant" (do correct me if I'm wrong), and then I tried to visualize the position function but I just got confused. I couldn't picture if it's linear or exponential.

So instead, I tried to solve for $v$ using $F = ma$.

$$ F = ma\\ F = m\frac{dv}{dt}\\ \frac{dv}{dt} = \frac{F}{m}\\ \int\frac{dv}{dt}dt = \frac{F}{m}\int dt\\ v = \frac{F}{m}t $$

which means the velocity graph is linear.

  • Am I doing things correctly?
  • What are the other ways of solving/thinking about this especially non-mathematical ones?

Thank you.

  • 1
    $\begingroup$ Your reasoning is correct. To visualize position, remember that velocity is a position derivative and do the same steps you did for acceleration. The graph will have a sudden change if the ball hits the ground in the end... Think about it. $\endgroup$ Jun 28, 2016 at 11:32
  • $\begingroup$ The force in Newton's second law is the net force and not just one of the forces acting on the moving object. $\endgroup$
    – nasu
    Sep 29, 2023 at 22:18

1 Answer 1


As the force $F$ is constant, and we know that the force of weight $mg$ is constant too; we can say the net force acting on the ball is constant whole the motion. But, from $t=0$ to $t=t_1$ the net force is $F-mg$ and after $t=t_1$ the net force is $-mg$. So, velocity graph will be linear but it has a breaking at $t=t_1$. For $0\le t\lt t_1$ the gradient of the graph is $\frac{F-mg}m$ and for $t\gt t_1$ the gradient of the graph is $-g$.

  • $\begingroup$ Importantly, the graph will likely also cross from positive to negative speed unless the ball hits the ground before coming to the peak of its trajectory. $\endgroup$ Sep 29, 2023 at 22:05

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