- 10

- 2

- Homework Statement
- Find the acceleration and displacement of a particle from its velocity versus time graph

- Homework Equations
- a=dv/dt

Please could I ask for help with the following question:

Part (a) is no problem. Acceleration is the gradient of the graph in regions OA and AB which gives 3 and 0.5

Part (b), I believe, requires me to calculate the greatest and least value of the gradient of the curve in region BC

Part (c), I believe, requires me to calculate the area under the whole graph.

My question is, can I solve parts (b) and (c) without working out the equation of the circle? Unless I have miscalculated, the equation of the circle is not particularly simple, the centre is not nicely located on integral values of x and y.

The answers the book gives are:

(b) (1/3)*(24 + 13*sqrt(3)) and (1/3)*(24 - 13*sqrt(3))

(c) 258

Thanks,

Mitch.

Part (a) is no problem. Acceleration is the gradient of the graph in regions OA and AB which gives 3 and 0.5

Part (b), I believe, requires me to calculate the greatest and least value of the gradient of the curve in region BC

Part (c), I believe, requires me to calculate the area under the whole graph.

My question is, can I solve parts (b) and (c) without working out the equation of the circle? Unless I have miscalculated, the equation of the circle is not particularly simple, the centre is not nicely located on integral values of x and y.

*My working was that the length of the chord is sqrt(15^2+20^2) = 25 and so to get the centre of the circle in order to find it's equation I should have to solve the following system:*

(x-15)^2+(y-20)^2 = 625

(x-30)^2+y^2=625

Once I had the equation of the circle I would differentiate it to find max min gradients and integrate it to find area under it.

is this the way to go?(x-15)^2+(y-20)^2 = 625

(x-30)^2+y^2=625

Once I had the equation of the circle I would differentiate it to find max min gradients and integrate it to find area under it.

is this the way to go?

The answers the book gives are:

(b) (1/3)*(24 + 13*sqrt(3)) and (1/3)*(24 - 13*sqrt(3))

(c) 258

Thanks,

Mitch.