Question

When you see a traffic light turn red, you apply the brakes until you come to a stop. Suppose your initial speed was 11.2 m/s, and you come to rest in 34.7 m. How much time does this take? Assume constant deceleration.

Answer 1

There are 4 basic 1D kinematics equations you can use under constant acceleration:\[x=x_0+v_0t+\frac{1}{2}at^2\]\[v=v_0+at\]\[x=x_0+\frac{1}{2}(v_0+v)t\]\[v^2=v_0^2+2a(x-x_0)\]These equations can be used to solve any 1D kinematics problem.

The information we are given is the initial speed (x_0) of 11.2 m/s and the displacement of 34.7 m. There are also other numbers hidden in there; we just need to look for them.

Since the car comes to rest at 34.7m past the point it started, that means we can define the initial x position as 0.

Additionally, since the car comes to a rest after this deceleration, we know that the final velocity is 0.

Now, if we look at the third equation:\[x=x_0+\frac{1}{2}(v_0+v)t\]We know all variables in this equation except t (which is the variable we are looking for). Perfect! Let's do some algebraic rearrangement before we plug values in:\[x=x_0+\frac{1}{2}(v_0+v)t \rightarrow x-x_0=\frac{1}{2}(v_0+v)t\]\[x-x_0=\frac{1}{2}(v_0+v)t \rightarrow \frac{2(x-x_0)}{(v_0+v)}=t\]Now we can go back and plug our values in:\[t=\frac{2(x-x_0)}{(v_0+v)}=\frac{2(34.7m-0m)}{(11.2\frac{m}{s}+0\frac{m}{s})}\]\[t=6.20s\]

Answer 2

Thank you so much!!!! =)