Question

A person driving her car at 43 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection (Fig. 2-29). The intersection is 15 m wide. Her car's maximum deceleration is -5.4 m/s2, whereas it can accelerate from 43 km/h to 70 km/h in 8.1 s. Ignore the length of her car and her reaction time.Figure 2-29If she hits the brakes, how far will she travel before stopping?
If she hits the gas instead, how far will she travel before the light turns red?m Should she try to stop, or should she speed up to cross the intersection before the light turns red?

Answer 1

Answer:

If she hits the brakes, she will travel 13m before stopping. If she hits the gas, she will travel 28 before the light turns red. She should try to stop

Explanation:

To know how far she will travel before stopping, we need to use a kinematic formula which initial final velocity (Vf), initial velocity (V0), acceleration (a) and distance traveled(x):

$V_{f}^{2}=V_{0} ^{2} +2ax$

The moment in which she stops is when the final velocity equals zero. In this case, initial velocity is 43km/h=12m/s, and its maximum deceleration is -5.4m/s^2. Plugging in these values and solving for x:

$0^2=(12m/s)^2+2(-5.4m/s)x\\x=13m$

She will travel 13m before stopping

If she hits the gas, we need another kinematic formula, which relates distance traveled, initial speed, time (t) and acceleration.

$x=v_{0}t+0.5a*t^2$

If we know her car can accelerate from 43km/h=12m/s to 70km/h=19m/s in 8.1 s, we can know its acceleration:

$a=\frac{19m/s-12m/s}{8.1s}=0.86m/s^2$

In this case, the time before the light turns red is 2.0s. Plugging in all those values:

$x=12m/s*2s+0.5*0.86m/s^2*(2s)^2=28m$

If she hits the gas, she will travel 28m before the light turns red. She won't even reach the intersection, so she would try to stop.