What is the deceleration of the vehicle?

[Kev1n]

1. I am looking to verify my calculation. Question A vehicle of mass 5Tonnes is traveling with a velocity of 80kmh-1 if it needs to stop within a dstnace of 0.5 km using a constant braking force what is the deceleration of the vehicle



2. V^2 = 2xSxA, A=V^2 / 2S



3. A=V^2/2s
80kmn = 22.222ms
A=22.222^2/2x500
A=493.8/1000
A=0.494
Deceleration = - 0.494

 

[Redbelly98]

Looks good

Welcome to PF.

 

[Kev1n]

Looks good

Welcome to PF.

appreciated thanks

 

[Kev1n]

Looks good

Welcome to PF.

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IET » Student and apprentice discussion forum » Calculation check Topic Title: Calculation check
Topic Summary: Deceleration
Created On: 18 October 2009 06:11 PM
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18 October 2009 06:11 PM


BK

Posts: 11
Joined: 28 September 2009

Can anyone give this the once over make sure I have the formula calcs etc right

Vehicle mass 5 Tonnes traveling with a velocity 80kmh. It needs to stop with in 0.5km using constant braking force

Deceleration:
(80kmh = 80x1000/3600 = 22.2m/s)

a=(v^2-u^2) / 2s
a=(0-22.2^2)/2x500
a=(493.83)/1000
a= 0.494ms
as Deceleration = -0.49

Just want to check I am on the right line
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19 October 2009 01:45 PM


tonysung

Posts: 579
Joined: 14 September 2001

It's a very good start. Perhaps you could add a few explanatory notes to your working so it can be checked properly.

E.g., "Deceleration80kmh = 80x1000/3600 = 22.2m/s)" should be 'velocity' or 'speed' in m/s.

Please have another go.

Kind regards

-------------------------
tony sung
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19 October 2009 07:11 PM


BK

Posts: 11
Joined: 28 September 2009

Apologies that should have been Velocity 22.2ms

My concern is whilst working out time taken

V/A=T, 22.2/0.4938 = 45s

I checked using a Velocity, Time distance calculator on the net and it calculates 22.5s

Therefore was I correct in using 2s in my initial calculation, it appears to be correct when I transformed the formula

 

[Redbelly98]

My concern is whilst working out time taken

V/A=T, 22.2/0.4938 = 45s

I checked using a Velocity, Time distance calculator on the net and it calculates 22.5s

Therefore was I correct in using 2s in my initial calculation, it appears to be correct when I transformed the formula

Yes, 2s was correct since the equation is

v² = 2 s a​

22.5 sec would be the time to go 500 m at a constant velocity (no deceleration) of 22.2 m/s. But in that case the vehicle would not come to a stop, as stated in the problem.