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## Homework Statement

You're speeding at 85 km/h when you notice that you're only 10m behind the car in front of you, which is moving at the legal speed limit of 60 km/h. You slam on your brakes, and your car decelerates at 4.2 m/s^2. Assuming the car in front of you continues at constant speed, will you collide? if so, at what relative speed? if not what will the distance between the cars at their closest approach?

**2. Homework Equations**

Equations of motion for constant acceleration

Please tell me if my reasoning is correct or where I am making a mistake.

I set the position where my car is when I start braking to be t=0, and this position to be x

My initial velocity is 85km/h=23.61 m/s

My acceleration is -4.2 m/s

For the car I may possibly collide with,

their initial velocity is 60 km/h= 16.6 m/sec

and their acceleration is zero

To come to a complete stop, my final velocity will be zero.

Using v=v

0=23.6 m/s -4.2m/s

and solving for t I get t=5.62 seconds to stop completely.

Next I tried to find my position in this time relative to where I started at x

using x=x

x=0+.5(23.6)(5.62)

=66.32 meters

Then I tried to see where car #2 would be in the same amount of time

using x=x

and I get x=10+16.67(5.62)+0

where last term goes to zero as car #2 is not accelerating.

x= 103.69 meters relative to the origin

Since the position of the two cars is not equal in 5.62 seconds, the two cars don't collide.

Equations of motion for constant acceleration

## The Attempt at a Solution

Please tell me if my reasoning is correct or where I am making a mistake.

I set the position where my car is when I start braking to be t=0, and this position to be x

_{0}=0My initial velocity is 85km/h=23.61 m/s

My acceleration is -4.2 m/s

^{2}For the car I may possibly collide with,

their initial velocity is 60 km/h= 16.6 m/sec

and their acceleration is zero

To come to a complete stop, my final velocity will be zero.

Using v=v

_{0}+at0=23.6 m/s -4.2m/s

^{2}tand solving for t I get t=5.62 seconds to stop completely.

Next I tried to find my position in this time relative to where I started at x

_{0}=0using x=x

_{0}+.5(v_{0}+v)tx=0+.5(23.6)(5.62)

=66.32 meters

Then I tried to see where car #2 would be in the same amount of time

using x=x

_{0}+v_{0}t+.5at^{2}and I get x=10+16.67(5.62)+0

where last term goes to zero as car #2 is not accelerating.

x= 103.69 meters relative to the origin

Since the position of the two cars is not equal in 5.62 seconds, the two cars don't collide.