You are driving your car along a country road at a speed of 21.5 m/s.

As you come over the crest of a hill, you notice a farm tractor 35.7 m
ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s.
You immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s2.
Will you hit the tractor before you stop? How far will you travel before you stop or collide with the tractor? If you stop, how far is the tractor in front of you when you finally stop?

Q1: distance car requires to stop? (m)

Q2: if you stop, how far is the tractor in front of you?

To determine the distance the car requires to stop, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s since the car stops)
u = initial velocity (21.5 m/s)
a = acceleration (-7.00 m/s^2)
s = distance traveled

Rearranging the equation, we have:

s = (v^2 - u^2) / (2a)

s = (0 - (21.5^2)) / (2 * -7.00)

s = (-462.25) / (-14.00)

s ≈ 33.02 m

Therefore, the car requires about 33.02 meters to stop.

To find the distance the tractor is in front of you when you finally stop, we need to determine how long it takes for the car to stop. We can use the equation:

v = u + at

Where:
v = final velocity (0 m/s)
u = initial velocity (21.5 m/s)
a = acceleration (-7.00 m/s^2)
t = time taken to stop

Rearranging the equation, we have:

t = (v - u) / a

t = (0 - 21.5) / (-7.00)

t ≈ 3.07 s

Now, to find the distance the tractor travels in this time, we multiply its velocity by the time:

distance = velocity * time

distance = 10.0 m/s * 3.07 s

distance ≈ 30.7 m

Therefore, when you finally stop, the tractor is approximately 30.7 meters in front of you.

To find the distance the car requires to stop, we need to determine the time it takes for the car to come to a stop. We can use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, since the car stops)
u = initial velocity (21.5 m/s)
a = acceleration (-7.00 m/s^2, assuming that deceleration is negative)
s = distance

Rearranging the equation:

0^2 = (21.5)^2 + 2(-7.00) * s

441.75 = 98s

Dividing both sides by 98:

s = 4.50 m

Therefore, the distance the car requires to stop is 4.50 meters.

If you stop, the tractor will continue to move forward for a certain distance. To find this distance, we need to calculate the time it takes for the car to come to a stop and then use that time to find how far the tractor would have traveled.

Again, using the kinematic equation:

v = u + at

Where:
v = final velocity (0 m/s, since the car stops)
u = initial velocity of the tractor (10.0 m/s)
a = acceleration of the car (-7.00 m/s^2)
t = time

Rearranging the equation:

0 = 10.0 - 7.00 * t

7.00t = 10.0

t = 10.0 / 7.00

t ≈ 1.43 s

Now, we can find the distance the tractor travels using:

s = ut + (1/2)at^2

s = 10.0 * 1.43 + (1/2) * (-7.00) * (1.43)^2

s ≈ 14.3 - 7.87

s ≈ 6.43 m

Therefore, if you stop, the tractor will be approximately 6.43 meters in front of you.

so 546 of y'all show this and not one of you did it ?