Question 1195353
A particle travelling in a straight line passes a fixed point O with a velocity of 16 m/s. Its
acceleration, a m/s^2, is given by a = 12- 6t, where t is the time in seconds after passing O. Calculate
a) the greatest speed attained by the particle in the original direction of the motion.
b) The distance from O when t = 2

<pre>
{{{a}}}{{{""=""}}}{{{12-6t}}}

{{{dv/dt}}}{{{""=""}}}{{{12-6t}}}

{{{dv}}}{{{""=""}}}{{{12*dt}}}{{{""-""}}}{{{6t*dt)}}}

{{{int(dv)}}}{{{""=""}}}{{{12*int(dt)}}}{{{""-""}}}{{{6*int(t*dt)}}}

{{{v}}}{{{""=""}}}{{{12t}}}{{{""-""}}}{{{3t^2}}}{{{"+"}}}{{{constant}}}

When t=0, the particle is at O, with velocity 16. So we substitute
t=0 and v=16

{{{16}}}{{{""=""}}}{{{12(0)}}}{{{""-""}}}{{{3(0)^2}}}{{{"+"}}}{{{constant}}}

So the constant is 16, and

{{{v}}}{{{""=""}}}{{{12t}}}{{{""-""}}}{{{3t^2}}}{{{"+"}}}{{{16}}}

Use the vertex formula or differentiating and setting the derivative = 0
to find that the maximum velocity is 

v=28 m/s when t=2 seconds.        <--answer to (a)   

To find the distance when t=2,

{{{v}}}{{{""=""}}}{{{12t}}}{{{""-""}}}{{{3t^2}}}{{{"+"}}}{{{16}}}

{{{ds/dt}}}{{{""=""}}}{{{12t}}}{{{""-""}}}{{{3t^2}}}{{{"+"}}}{{{16}}}

{{{ds}}}{{{""=""}}}{{{12t*dt}}}{{{""-""}}}{{{3t^2*dt}}}{{{"+"}}}{{{16*dt}}}

{{{int(ds)}}}{{{""=""}}}{{{12int(t*dt)}}}{{{""-""}}}{{{3int(t^2*dt)}}}{{{"+"}}}{{{16int(dt)}}}


{{{s}}}{{{""=""}}}{{{6t^2}}}{{{""-""}}}{{{t^3)}}}{{{"+"}}}{{{16t}}}{{{""+""}}}{{{constant}}}

When t=0, the particle is at O, so at that instant the distance s=0, So
we substitute t=0 and s=0

{{{0}}}{{{""=""}}}{{{6(0)^2}}}{{{""-""}}}{{{(0)^3)}}}{{{"+"}}}{{{16(0)}}}{{{""+""}}}{{{constant}}}

So the constant is 0

{{{s}}}{{{""=""}}}{{{6t^2}}}{{{""-""}}}{{{t^3)}}}{{{"+"}}}{{{16t}}}

We substitute t=2

{{{s}}}{{{""=""}}}{{{6(2)^2}}}{{{""-""}}}{{{(2)^3)}}}{{{"+"}}}{{{16(2)}}}

s = 48 meters.       <--answer to (b)

Edwin</pre>