Question

At time t seconds after starting the position coordinate x cm of a point moving in a straight line XOX' is given by x=4+4t-3t^2. a.) Prove that the acceleration is constant and find its value. b.) find when the pint is at 0. c.) Find the position and velocity when the elapsed time is 5 second. d.) Find the volocity when (i) x=4, (ii) x=-11.

Answer 1

if x=4+4t-3t^2
then V=4-6t
and A=-6
Can you take it from this point?

Answer 2

you can double differentiate the equation for position of the particle to get the acceleration and by double differentiating this equation, you get acceleration=-6 as munkeyman1985 says
next to find that at what instant the pint is at 0, simply put x=0
so we get 4+4t-3t^2=0
so by solving the quadratic equation, we get t=2 or -2/3
we reject the value-2/3 as time cannot be negative
the next part of the question can be solved by putting the values of x and t in the equations as given by munkeyman
i hope this helps you:)

Answer 3

Thx you helped me a lot. :)

Answer 4

thanks for the medal