Question

When traveling with a velocity of 15m.s ^-1 a cyclist suddenly realizes that he has to stop at an intersection 20m ahead. By apply breaks he can slow down at 2m.s^-2. Takes him 3s to reach intersection. Explain by using calculations if he is able to stop in time.

Answer 1

We can find the stopping distance of cyclist from the kinematic equation:

$$d=v_0t+\dfrac{1}{2}at^2,$$

here, $d=20\ m$ is the stopping distance of cyclist, $v_0=15\ \dfrac{m}{s}$ is the initial velocity of cyclist, $t$ is time that cyclist takes to finally stop, $a=-2.0 \dfrac{m}{s^2}$ is the deceleration of cyclist.

Let's substitute $d$, $v_0$ and $a$ into the equation, we get:

$$20 = 15t + 0.5\cdot(-2)t^2,$$ $$t^2-15t+20=0.$$

Let's solve this equation for $t$. This quadratic equation has two roots:

$$t_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{15+\sqrt{(15)^2-80}}{2}=13.52\ s,$$ $$t_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}=\dfrac{15-\sqrt{(15)^2-80}}{2}=1.48\ s.$$

So, the cyclist takes $t=1.48\ s$ to finally stop (we don't need the first root because the second one already satisfies the equation). As we can see from the calculations, it takes him 3 s to reach the intersection, but in fact, he finally stop in 1.48 s. Therefore, he is able to stop in time.

Answer:

The cyclist is able to stop in time.