Question

Carol is driving at 25 km/h when she sees a stop light turn yellow. If it takes her 1.2 s to begin braking, and the car's acceleration is a constant -4.0 m/s2, what is the total distance she travels from the time she first sees the light turn until the car stops?

Answer 1

Answer:

The total distance she travels from the time she first sees the light turn until the car stops is 6.02 meters

Step-by-step explanation:

The car when Carol begins to hit the brakes is traveling with constant deceleration, that means we have to use uniformly accelerated motion equations to solve for the distance traveled, but first we need to analyze the given data.

Given information

• Initial velocity,  $v_i = 25 km/h$
• Final velocity,  $v_f = 0$
• Time, t = 1.2 s
• Acceleration, $a= -4.0 \cfrac{m}{s^2}$

Almost all information is on standard units except the initial velocity, so we can convert it to meters per second.

$v_i = 25 \cfrac{km}{h} \times \cfrac{1h}{3600s}\times \cfrac{1000m}{1km}\\v_i=6.94 \cfrac ms$

Also notice that the final velocity is 0 m/s, since the car stops at the end.

Setting up the motion equation.

Since we have constant acceleration and since the goal is to find the distance we can use the following equation for uniformly accelerated motion.

$v_f^2 = v_i^2 +2ad$

Here d is the distance traveled, replacing the given information we get

$0=\left(6.94 \cfrac ms\right)^2 +2\left( -4 \cfrac{m}{s^2}\right)d$

Solving for d

We can solve for the distance traveled d by moving every other number to the other side.

$-\left(6.94 \cfrac ms\right)^2 =2\left( -4 \cfrac{m}{s^2}\right)d$

We can divide both sides by -8

$\cfrac{6.94^2}{8} m= d$

So we get

$\boxed{d=6.02 m}$

The car has traveled 6.02 meters before stopping.