Problem with Applications of Calculus (1 Viewer)

[aurel]

I have no idea how to do the graphical part of this topic. For example, in questions where they give you the displacement graph of an object and ask you to skectch the graphs for velocity and acceleration. Help please! Thanks!

 

[word.]

basically all you need to know is that v = dx/dt, a = dv/dt

the velocity-time graph is the derivative of the displacement-time graph

so every value in your velocity curve corresponds to the gradient of the displacement curve for any given t.

as a result,
when the displacement curve has a stationary point (turning point, inflexion), the gradient is 0 hence this is where the velocity curve crosses the x-axis.

when the displacement curve has a positive gradient (rising from left to right) then the velocity curve is above the x-axis

the higher the velocity curve is, the steeper the gradient of the displacement curve at that point

etc.

works similarly for acceleration

 

[aurel]

Hey word. thanks for the reply. I understand the part about the derivative, and about how the velocity is zero at the stationary points. i now understand the velocity part, but not how to get the acceleration graph from velocity.

So the stationary point on the velocity graph is where he acceleration curve crosses the x-axis. But, how do i know which part of the line is above or below the axis? For example, when the velocity graph is a parabola, concave down?

Thanks!

 

[word.]

since acceleration is the rate of change of velocity, the values of the acceleration curve correspond to the gradient of the velocity curve at a given t

so when the velocity curve has a turning point, the acceleration curve crosses the x-axis(time axis). it's better to understand why in your head rather than to memorise it all.

take for example in this scenario, velocity has a maximum turning point. this means the speed of the moving object reaches a top speed and doesn't go any faster than that speed, i.e. the acceleration of the object is 0 at the top speed (i.e. the curve crosses the x-axis), and then since the velocity drops after the max turn point, we're slowing down i.e. decelerating, so the acceleration is negative.

 velocity
   |   _        /     
   |  / \      /
   |------------>time           
   |/     \__/
   |

the curve has an increasing gradient, so the acceleration curve will be above the time-axis, but then it his a maximum turning point, so the acceleration is 0 at that point.

since the gradient of the velocity curve is positive as it reaches the turning point, the acceleration curve is positive as it reaches the time-axis.

            /     \
 positive  /       \  negative
 gradient /         \ gradient

now between the max turn point and the min turn point, the gradient is negative, so the acceleration curve is below the time-axis until it starts rising again.

note the inflexion point in the middle of the velocity curve, this is a change in the concavity of the velocity curve, and indicates an inflexion point in the acceleration curve.

 acceleration
   |\        /     
   | \      /
   |------------>time           
   |   \__/
   |

a good thing to try is to graph functions f(x) and its derivative f'(x)
and observe the relationships between them. then if you assume f(x) is the displacement-time graph, f'(x) will be its velocity graph, or f(x) = velocity, f'(x) = acceleration

 

[aurel]

Thanks so much for your help word. i think i finally get it! yay! i have a maths exam on this topic tomorow, so hopefully the understanding shall be put into correct practice. Thanks again for your time and help!