Question

Picture attached

Answer 1

So, for the 1st, it appears they give you an equation for that point :
\[a=32-80t^2\] they are asking for a at that point t=0.3 so that gives us
\[a=32-80(0.3)^2\] for the second point t=0.2 look at the box they provided for the point

Answer 2

position means distance and not acceleration, right?

Answer 3

yes

Answer 4

i actually misread this, you will need to use all the bits they give you to determine how much displacement has occurred piecewise

Answer 5

and by the way, the equation for the straight line is a = 32- 80t (it's obviously must be in the first degree)

Answer 6

Can you help me out? I've done lots of integrals and but still not sure on accuracy of my answer.

Answer 7

So, despite it saying it's a line, i would use the eqn they give. for the first bit, we would have to use two integrals:
\[\int_0^{0.2} 24-200t^2 dt +\int_{0.2}^{0.3}32-80t^2dt \] would be the total displacement from 0->0.3 for the second bit, it would just be the first integral to find the displacement

Answer 8

actually these are accelerations, so that would give velocity... its been a while since i did physics lol

Answer 9

but getting the first integral of acceleration function will give us velocity

Answer 10

yeah, you would then take the integrals of velocity to determine the displacement

Answer 11

those will give me values not a function, so i don't think if that's really applicable

Answer 12

\[\int24-100t^2dt=v(t)=24t-\frac{100t^3}3\\\int_{0}^{0.2}24t-\frac{100t^3}3dt=x(t)-x(0)\]

Answer 13

would be the second part

Answer 14

@ganeshie8 , can you?