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IMPOSSIBLE isaac physics question

The speed of motion of a trolley before and after left (1.7 ± 0.1)s of acceleration is measured. Before the acceleration, the trolley moved left bracket (100 ± 1)mm in left bracket, (1.78 ± 0.01)s; after the acceleration it moved the same distance in (0.74 ± 0.01)s. Calculate the relative uncertainty of the measured acceleration.

HELP Please I am doing a = v/t but first I find uncertainties in each
In first velocity uncertainty is 1% + 0.56% = 1.56% (rounded but not in final answer)
In second uncertainty is 1% + 1.35% = 2.35%
So if I do (absolute uncertainty)/(velocity) for both of them I get the absolute velocities. So delta(v) = v2-v1=56.2-135. Then I add the absolute velocities together and find the percentage uncertainty. Then I add the uncertainty in delta(v) to the percentage uncertainty of the total time and I get 7.0338% but when I give this to 2sf ISAAC SAYS ITS WRONG! I've tried it three times already I get the same answer please show where i'm wrong or this question must be impossible
Reply 1
Original post by isaacphysicshlp
The speed of motion of a trolley before and after left (1.7 ± 0.1)s of acceleration is measured. Before the acceleration, the trolley moved left bracket (100 ± 1)mm in left bracket, (1.78 ± 0.01)s; after the acceleration it moved the same distance in (0.74 ± 0.01)s. Calculate the relative uncertainty of the measured acceleration.

HELP Please I am doing a = v/t but first I find uncertainties in each
In first velocity uncertainty is 1% + 0.56% = 1.56% (rounded but not in final answer)
In second uncertainty is 1% + 1.35% = 2.35%
So if I do (absolute uncertainty)/(velocity) for both of them I get the absolute velocities. So delta(v) = v2-v1=56.2-135. Then I add the absolute velocities together and find the percentage uncertainty. Then I add the uncertainty in delta(v) to the percentage uncertainty of the total time and I get 7.0338% but when I give this to 2sf ISAAC SAYS ITS WRONG! I've tried it three times already I get the same answer please show where i'm wrong or this question must be impossible

Take a look at this answer
https://www.thestudentroom.co.uk/showthread.php?t=5033458

The answer is wrong you put it in the website and it says its wrong.
Reply 3
Original post by isaacphysicshlp
The answer is wrong you put it in the website and it says its wrong.

Could you send the question link please?
The answer is 10i've done this question for 30 mins
Reply 5
Original post by UMMMM123
The answer is 10i've done this question for 30 mins


How did you get this answer? Thanks
Original post by lukem06
How did you get this answer? Thanks


Hi, I understand that you are new to TSR. Please avoid posting in thread that is more than 1 year ago.
I would try to outline and “explain” the working for the question.

The objective is to find Δaa \frac{\Delta a}{a} . It is given by
Δaa=Δ(Δv)Δv+Δtata \dfrac{\Delta a}{a}=\dfrac{\Delta \left( \Delta v \right)}{\Delta v}+\dfrac{\Delta {{t}_{a}}}{{{t}_{a}}}

where Δv is the change in velocity, Δ(Δv) is the absolute uncertainty of change in velocity, and ta is 1.7 s.

Initial velocity is found using (100 ± 1) mm in (1.78 ± 0.01) s and also use them to find the associated uncertainty.

The final velocity and its uncertainty are found using (100 ± 1) mm in (0.74 ± 0.01) s.

The change in velocity is the final velocity minus the initial velocity.

The absolute uncertainty for the change in velocity is found using the following rule:
When a physical quantity A is found using by
A = B - C
with the absolute uncertainty of B and C to be ΔB and ΔC, respectively. The absolute uncertainty of A is
ΔA = ΔB + ΔC

Then use the uncertainty of the change in velocity and (1.7 ± 0.1) s to find
Δaa=Δ(Δv)Δv+Δtata \dfrac{\Delta a}{a}=\dfrac{\Delta \left( \Delta v \right)}{\Delta v}+\dfrac{\Delta {{t}_{a}}}{{{t}_{a}}}

If you still have issues after going through it, start a new thread and describe your problem(s), as the thread would be locked soon.

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