# Can some one check this over? (1 Viewer)

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#### Eddard

Hey... I need help on a question i know how to calculate the decceleration of the truck in the problem below but im not completly sure about how you would be able to find the conclusion to this question? Help would be appreiated A 920 kg car is 65m from a concrete barrier travelling at 125 km/h, when the driver notices and slams on the breaks. The fritional force acting on the car is 8600N. Is the driver able to stop the car in time? so this is what i got:

M= 920kg
D=65m
V1=125 km/h
F(friction)=8600N
A=?
v1=0

F(net)=ma
-F(friction)=ma
A=-F(fric..)
=-0.935 m/s ^2

V2^2=V1^2 + 2a*∆d

Rearranged the equation:
∆d= V2^2/2a∆d - square of V1

Then when i sub in the numbers and solve the equation I conclude that the car stops 2.11 meters away from the concrete barrier....but im not sure if i did this right and if any part of this equation doesnt make sense ill try to explain....

Last edited:

#### Andrew Mason

Homework Helper
Hey... I need help on a question i know how to calculate the decceleration of the truck in the problem below but im not completly sure about how you would be able to find the conclusion to this question? Help would be appreiated A 920 kg car is 65m from a concrete barrier travelling at 125 km/h, when the driver notices and slams on the breaks. The fritional force acting on the car is 8600N. Is the driver able to stop the car in time? I am not sure what you are doing here. The easiest way to do this is to use energy. The force x stopping distance must equal the initial kinetic energy of the truck. If the stopping distance is greater than 65 m then it hits the wall.

Since Fd = KE, the stopping distance is: d = KE/F

Work that out and see if d is less than 65 m.

AM

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