With what speed does the car then strike the tree?

In summary, the car slams on the brakes when it sees a tree blocking the road. The car slows uniformly with acceleration 5.10m/s^2 for 4.40s, making straight skid marks 64.6 m long ending at the tree. With what speed does the car then strike the tree?
  • #1
chocolatelover
239
0

Homework Statement


THe driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration 5.10m/s^2 for 4.40s, making straight skid marks 64.6 m long ending at the tree. With what speed does the car then strike the tree?

Homework Equations


vxf=vxi+axt
d=(vi+vf/2)t


The Attempt at a Solution



a=5.10m/s^2
t=4.40s

- - - - -
a=5.10m/s^2
_________________________________
vi=0 4.40s

d=64.6m

d=(vi+vf/2)t
64.6m=(om/s+vf/2)4.40

Then I would solve for vf, right?

Thank you very much
 
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  • #2
I'm having trouble deciphering what you've done. You need to find the initial speed first, then you can find the final speed.
 
  • #3
Thank you

Wouldn't vi be 0? If it isn't 0, then I could't determine the initial speed and the final speed if I'm not given either one, right? because I could use the equation vxf=vxi+axt, but if I don't know what vf or vi are, then I can't solve this, right?

Is this completely wrong? if so, could you please give me a hint?

Thank you very much
 
  • #4
what you're trying to solve for is the final velocity. If the initial velocity was zero then that wouldn't make sense since the driver wouldn't have to brake at all. What you need to do is find the initial velocity from the information given before finding the final velocity. Think of a kinematic equation involving intial velocity, acceleration, distance and time.
 
  • #5
I know that xf=xi+vxit+1/2axt^2 Does this look right?

64.6m=0m+vxi(4.40s)+1/2(5.10m/s^2)(4.40s^2)
vxi=3.462

vxf=vxi+axt

vxf=3.462m/s+5.10m/s^2(4.40s)
=25.902m/s

Thank you very much
 
  • #6
The initial speed will not be 3.462m/s. Remember to be careful with signs.
 
  • #7
64.6m=0m+vxi(4.40s)+1/2(5.10m/s^2)(4.40s^2)

I can't seem to find what I'm doing wrong. Xf is 64.6m and xi is 0m, right?

Thank you
 
Last edited:
  • #8
Well if you're slowing down the acceleration will be in the opposite direction to the direction of travel and it will therefore be negative.
 
  • #9
How long would a road have to go unused in order for someone to find it blocked by a tree?

This is the dumbest physics problem evar!
 
  • #10
Would this work?

x=64m
a=-5.10m/s
t=4.4s

x=vit+1/2at^2

64m=vi(4.4s)+1/2(-5.1m/s)(4.4s)^2

vi=25.8m/s

vf^2=vi^2+2ax

vf^2=25.8m/s^2+2(-5.10m/s^2)(64m)

=3.6m/s

Thank you
 
  • #11
chocolatelover said:
Would this work?

x=64m
a=-5.10m/s
t=4.4s

x=vit+1/2at^2

64m=vi(4.4s)+1/2(-5.1m/s)(4.4s)^2

vi=25.8m/s

vf^2=vi^2+2ax

vf^2=25.8m/s^2+2(-5.10m/s^2)(64m)

=3.6m/s

Thank you

That looks ok to me.
 
  • #12
Thank you very much

Regards
 

1. What is the definition of speed in this context?

Speed is the rate at which an object moves or changes position over time. In this context, it refers to how fast the car is traveling.

2. How is speed measured?

Speed is typically measured in units of distance per time, such as miles per hour or meters per second. In this case, it would be the distance the car travels before hitting the tree divided by the time it takes to reach the tree.

3. Does the speed of the car affect the impact with the tree?

Yes, the speed of the car will impact the force of impact with the tree. The higher the speed, the greater the force upon impact.

4. Is there a formula for calculating the speed of the car?

Yes, the formula for calculating speed is distance divided by time (s = d/t). In this case, the distance would be the distance the car travels before hitting the tree and the time would be the time it takes to reach the tree.

5. Can the speed of the car be changed before hitting the tree?

Yes, the speed of the car can be changed before hitting the tree. This can be done by either accelerating or decelerating the car, or by changing the direction of the car's movement.

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