The Student Room Group

Deceleration

Ok out of all the equations and equation pyramids I can remember - I can't think of one to apply to this question on deceleration...

Q) The part of the space shuttle (orbiter) which returns to Earth has a mass of 78,000Kg and lands at a speed of 100m/s. After touchdown it takes 50 seconds to decelerate and come to a halt.

a) Calculate the deceleration of the orbiter
b) Calculate the force needed to bring the orbiter to a halt.

I know how to do b), using Force = Acceleration x Mass, except I'm not sure how to calculate the deceleration in part A, any ideas anyone?

Edit: How about Change in velocity = acceleration x time? so er.. that would be Deceleration = Change in velocity / time = 100 / 50 = 2m/s???

Edit:: So then Force= Acceleration x Mass... 2 x 78,000 = 156,000 .. or would you divide if its DEceleration?? Hellllllllllp
Your answers are RIGHT!!

...except that acceleration is 2 m/s2 or ms-2 if you prefer.
CrazyChemist
Your answers are RIGHT!!

...except that acceleration is 2 m/s2 or ms-2 if you prefer.


But - how can it be 2m/s^2 if its decelerating?? surely that would be acceleration
It asks for the deceleration - the deceleration is 2ms-2. There's no minus sign before the 2 because it is already clear that it is decelerating not accelerating.
CrazyChemist
It asks for the deceleration - the deceleration is 2ms-2. There's no minus sign before the 2 because it is already clear that it is decelerating not accelerating.


So I would use -2 all the time if I'm talking about deceleration and just the regular 2 for accelaration?
Reply 5
No the unit of acceleration is ms^-2 this never changes! Acceleration is a vector quantity meaning that it must have direction - so if acceleration is positive (forwards in direction) then deceleration (backwards in direction) must be negative.
Trolley
No the unit of acceleration is ms^-2 this never changes! Acceleration is a vector quantity meaning that it must have direction - so if acceleration is positive (forwards in direction) then deceleration (backwards in direction) must be negative.


So what's the unit of deceleration??
hearthethunder
So I would use -2 all the time if I'm talking about deceleration and just the regular 2 for accelaration?
nono - wrong end of stick!

The units ms-2 and m/s-2 are the same thing. There's no such unit that i know of "ms2".

I thought you were reffering to the fact that deceleration is usually written as negative acceleration, eg. -2ms-2. My post was saying that since it is already stated that it is deceleration, there is no need for the "-" in front of the 2: this one >2ms-2 not 2ms- this one>2

I realise it was unclear - apologies.
CrazyChemist
nono - wrong end of stick!

The units ms-2 and m/s-2 are the same thing. There's no such unit that i know of "ms2".

I thought you were reffering to the fact that deceleration is usually written as negative acceleration, eg. -2ms-2. My post was saying that since it is already stated that it is deceleration, there is no need for the "-" in front of the 2: this one >2ms-2 not 2ms- this one>2

I realise it was unclear - apologies.


Ohh so use the same units for both of them yeah I get it now.. it's just in class we've been taught that acceleration is m/s2 rather than m/s-2.
hearthethunder
Ohh so use the same units for both of them yeah I get it now.. it's just in class we've been taught that acceleration is m/s2 rather than m/s-2.
Dammit - there's still a little point here:

unit of acceleration and deceleration are the same (echoing trolley)

This unit can be written as

ms-2

OR

m/s2

Absolutely and categorically NOT

m/s-2
CrazyChemist
Dammit - there's still a little point here:

unit of acceleration and deceleration are the same (echoing trolley)

This unit can be written as

ms-2

OR

m/s2

Absolutely and categorically NOT

m/s-2


Aha!!! *moment of realisation* ... *cements into mind*
hearthethunder
Aha!!! *moment of realisation* ... *cements into mind*
:biggrin: great
Reply 12
Poor Chemist...hehe