znx
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A truck of mass 2300 kg is travelling at a constant speed of 22 m/s along a dry, level road. The driver reacts to a hazard ahead and applies the brakes to stop the truck. The reaction time of the driver is 0.97 s. The brakes exert a constant braking force of 8700 N.

1. Calculate the magnitude of the deceleration of he truck when braking.
I tried doing it two different ways,
1st was to use F = ma and I did 8700/2300 to get 3.78
2nd was to use a = (final v - initial v)/t, to get a = -22/0.97 = 22.7

I could not find a mark scheme to this question, so I do not know which method, if any is correct.
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Callicious
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(Original post by znx)
A truck of mass 2300 kg is travelling at a constant speed of 22 m/s along a dry, level road. The driver reacts to a hazard ahead and applies the brakes to stop the truck. The reaction time of the driver is 0.97 s. The brakes exert a constant braking force of 8700 N.

1. Calculate the magnitude of the deceleration of he truck when braking.
I tried doing it two different ways,
1st was to use F = ma and I did 8700/2300 to get 3.78
2nd was to use a = (final v - initial v)/t, to get a = -22/0.97 = 22.7

I could not find a mark scheme to this question, so I do not know which method, if any is correct.
0.97 is the reaction time: he won't be decelerating during this time. I imagine that is for another part of the question? Or is there no other part? ;-;
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Sajjad_K9
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Yeah so where's the stopping time?
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znx
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(Original post by Callicious)
0.97 is the reaction time: he won't be decelerating during this time. I imagine that is for another part of the question? Or is there no other part? ;-;
The only other part of the question was:
Show that the stopping distance of the truck is about 85m
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Callicious
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(Original post by znx)
The only other part of the question was:
Show that the stopping distance of the truck is about 85m
Ah, then that'll be what the reaction time/etc is for.

Reaction time * initial velocity + Deceleration Distance = Stopping Distance

(probably... *checking* yes, yes it is) (WAIT NOT THAT I USED THE ANSWER AND PROVED IT YOU SHOULD DO THAT YOURSELF TO PROVE IT)
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3pointonefour
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Just use F = ma
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znx
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(Original post by Callicious)
Ah, then that'll be what the reaction time/etc is for.

Reaction time * initial velocity + Deceleration Distance = Stopping Distance

(probably... *checking* yes, yes it is) (WAIT NOT THAT I USED THE ANSWER AND PROVED IT YOU SHOULD DO THAT YOURSELF TO PROVE IT)
Thanks, I realised that I was finding the deceleration using the reaction time. I managed to get the right answer in the end
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