# Why acceleration is positive in this graph?

Please explain this graph to me as why acceleration is positive

• There is a rule that tells us the function is convex if its second derivative is greater than zero. $a=\dfrac{d^2x}{dt^2}$ and it is greater than zero, so $x$ is convex. Another explanation: the acceleration is positive, then the velocity $v(t)$ is an ascending function; we could consider an elementary case, when $V=ax$, i.e. $a$ is constant. Then $x\sim Ca^2, C>0$, so the path would represent a part of parabola with "branches up". – nicael Apr 13 '18 at 9:40
• @nicael I think your answer is fine and if I were you I wouldn't have deleted it. If you would like to undelete it I think it would make a useful contribution. This is an elementary question but I still think it's a valid question. We were all beginners at some point. – John Rennie Apr 13 '18 at 10:41
• @John well, ok! – nicael Apr 13 '18 at 13:38

The velocity is the gradient of the line on your position:time graph. If I roughly estimate the gradient by eye I get something like this:

And the acceleration is the gradient of the velocity:time graph

• Alternatively, the velocity is increasing (becoming less negative) with time, so the acceleration must be positive. – jim Apr 13 '18 at 9:46
• Please tell me why velocity is increasing? – brahamdeep singh Apr 13 '18 at 11:19
• @brahamdeepsingh The velocity starts out as a negative number and it becomes less negative. That means the change in the velocity is a positive number i.e. the velocity is increasing. – John Rennie Apr 13 '18 at 11:21

There is a rule that tells us the function is convex if its second derivative is greater than zero.

$a=\dfrac{d^2x}{dt^2}$ and it is greater than zero, so $x$ is convex.

As another example, you could consider that:

• the acceleration is positive
• then the velocity $v(t)$ is an ascending function
• we could consider an elementary case, when $V=ax$, i.e. $a$ is constant.
• then $x\sim Ca^2, C>0$, so the path would represent a part parabola with "branches up"