Polynomials 2-Maths

Question

An object moves in a straight line so that it's distance, y cm from a fixed point O on the line is given by the equation y=t² -12t +36,where t is the time inow seconds after it starts moving. Deduce and write down the minimum distance between the object and O. Also, write down the time that corresponds to the minimum distance. Finally, describe the motion of the object after the 6th second.

Answer 1

You have to notice that the closest distance to the fixed point will be when the objects stops moving in the negative direction. dy/dt = V_f = 0

(unless object actually goes past the fixed point before it stops) then closest distance =0)

a. If you know calculus, then you must find when the derivative of the equation for distance equals zero...

distance equation: y= t² -12t +36

derivative of distance equation dy/dt = 2t - 12, set to 0=2t-12,

so time of closest distance is when t= 6s

Now you can go back to the above equations and use 6s to find out where it is (distance) and what its doing (dy/dt)

b.If you DON"T know calculus, then you'd have to notice that y= t² - 12t +36 is the displacement equation for constant acceleration " d=v_it + 1/2 at² + d_i " with

a = 2 cm/s, v_i = -12 cm/s, and d_i = 36cm

Then use that data and the defining equation for acceleration " a = (v_f - v_i)/t " to

find the time when v_f =0. now use that time in each equation to find where it is and what its doing.