# Ambiguous acceleration ques!Watch

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#1
Hello.. got a maths question regarding acceleration. The question isn't difficult, but i'm not sure about it because of acceleration being a vector quantity. Look at the attachment - apologies that the diagram is so poor, but it is a displacement time graph for a car. Ques: What are the positions of the car when the acceleration is a)Negative b)Zero c)Positive
I can see that the acceleration would be positive between O and A and between B and C it's negative, but what about after the car has stopped? Is the acceleration actually classed as negative when it is speeding up, bewteen B and C because we take the acceleration to be positive in the direction with increasing displacement? Or am i confusing things too much?

Cheers...
0
14 years ago
#2
(Original post by Callum10)
Hello.. got a maths question regarding acceleration. The question isn't difficult, but i'm not sure about it because of acceleration being a vector quantity. Look at the attachment - apologies that the diagram is so poor, but it is a displacement time graph for a car. Ques: What are the positions of the car when the acceleration is a)Negative b)Zero c)Positive
I can see that the acceleration would be positive between O and A and between B and C it's negative, but what about after the car has stopped? Is the acceleration actually classed as negative when it is speeding up, bewteen B and C because we take the acceleration to be positive in the direction with increasing displacement? Or am i confusing things too much?

Cheers...
even though i have been competing in a fast ass speed skate i'll help you out

it has negative acc.( as in it is decelerating ) at points B to C

it has zero acc. at the peak and the at the beginning and end

positive acc. A through to B

the GRADIENT of the d-t graph gives the velocity

and just so you know the gradient of the v-t graph gives the acceleration
#3
i dont think it's decelerating from b to c actually....it's getting faster.
0
14 years ago
#4
accln is given by
a = d²x/dt²

now dx/dt is just the slope of that curve, and d²x/dt² is the rate at which the slope is changing.
When d²x/dt² is +ve, then that's a minimum. i.e. the curve is concave downwards
When d²x/dt² is -ve, then that's a maximum. i.e. the curve is concave upwards
When d²x/dt² is zero, then that's a point of inflection.

from O to A is concave downwards => "minimum"-ish => d²x/dt² is positive
from A to C is concave upwards => "maximum"-ish => d²x/dt² is negative
from C to D is concave downwards => "minimum"-ish => d²x/dt² is positive

A and C look like points of inflexion, so d²x/dt² = 0

Analysing the curve.
Let the car start at O with what looks like zero velocity (since the slope looks like zero)
From O to A is +ve for d²x/dt² so the car is accelerated and its velocity increases as we can see by the increase of the slope of the x-t curve.
Point A is a point of inflection, so zero accln.
From A to B the curve is concave upwards so accln is -ve and the car slows down.
The slope is zero at B, and this is where the car has stopped - it has zero velocity.
After B, from B to C the accln is still -ve and the car now moves backwards, its velocity increasing (numerically) all the time, until point C is reached.
Now the accln is +ve and the car's backwards velocity is reduced until at point D when it is at rest again.
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