# Considering velocity-distance graph [closed]

A car is travelling at constant velocity. Its brakes are then applied, causing uniform deceleration. Which graph shows the variation with distance s of the velocity v of the car?

The solution is A, but I don't know how to solve this problem because it is a velocity-distance graph. I only know how to draw basic graphs like V-t and S-t graphs as shown below. Could you explain the work solution to me please? Thank you very much.

• $v_f^2=v_i^2+2as$ is the constant acceleration kinematic equation to consider. Commented Jun 3, 2016 at 7:55

Consider the graph only from the moment when the brakes are applied, that's the time $t=0$ here. The uniform deceleration means that the distance travelled after that is $$s = v_0 t - \frac 12 a t^2$$ while the velocity is $$v = v_0 - at$$ We want to draw $v(s)$ as a function of $s$ but the second formula depends on $t$. So we have to find $t$ from the first equation. It's a quadratic equation and the solution is $$t = A + B \sqrt{C-s}$$ for some constants $A,B,C$. Substitute it to get $$v = E + F \sqrt{C-s}$$ We know that the velocity is a continuous function of $s$ and it decreases after the critical moment. But the character of the decrease is $F\sqrt{C-s}$, as we got, so the graph must look like the left-right-reflected graph of $y=\sqrt{x}$ near the point where the car stops, $s=C$, and that's clearly the graph (A). You should be able to calculate the coefficients $A,B,C,D,E,F$ easily.