With what speed does the car then strike the tree?

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The discussion revolves around calculating the speed of a car just before it strikes a tree after braking. The car decelerates uniformly at 5.10 m/s² over 4.40 seconds, covering a distance of 64.6 meters. Participants clarify that the initial speed must be determined first, as assuming it to be zero is incorrect. The correct approach involves using kinematic equations to find the initial velocity, which is calculated to be approximately 25.8 m/s, leading to a final speed of about 3.6 m/s upon impact. The conversation emphasizes understanding the signs of acceleration and the importance of correctly applying physics equations.
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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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I'm having trouble deciphering what you've done. You need to find the initial speed first, then you can find the final speed.
 
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
 
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.
 
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
 
The initial speed will not be 3.462m/s. Remember to be careful with signs.
 
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:
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.
 
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
 

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