A motorist is driving at 20 m/s when she sees that a traffic light 200 m ahead has just turned red. She knows that this light stays red for 15 s, and she wants to reach the light just as it turns green again. It takes her 1.0 s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?

The initial velocity of the motorist is $v_i=20m/s$

The distance of the red light from the bike is $S=200m$

The reaction time to apply break is$t=1sec$

The traffic light stays red for 15 secs

The distance traveled by a motorist in reaction time is

$$\begin{gathered} s_1=v_i\left(t\right)=20\left(1\right) \\ {s}_1=20m \end{gathered}$$

The breaks are applied after covering the distance of 20 m.
Thus the distance remained between the bike and the red light is localid="1648463946152" $s_2=200-20=180m$

Also, the time remained for the bike to reach the red light is$t=(15-1)=14sec$

Now, write the equation of motion,
localid="1648463978194" $$\begin{gathered} s_2=ut+\frac{1}{2}a{t}^2 \\ 180=20\left(14\right)+\frac{1}{2}a{\left(14\right)}^2 \\ 98a=-100 \\ a=-1.02m/s^2 \end{gathered}$$

Thus, the acceleration of the bike is -1.02 m/s²

Now, write the equation of motion to determine the final velocity of the bike when the light turns green

role="math" localid="1648463927622" $$\begin{gathered} v_f=v_1+at \\ {v}_f=20+(-1.02)\left(14\right) \\ {v}_f=20-14.28 \\ {v}_f=5.72m/s \end{gathered}$$
Thus, the final speed of the bike is 5.72 m/sec