Question 364182
The height (as a function of time, t) of an object falling from 100 meters is described by:
{{{h(t) = 100-16t^2}}}
i) Find the velocity at t = 2
The velocity is the rate of change of distance with respect to time, t.
If we take the first derivative of the height fuction, we will get the eqation for the velocity.
{{{dh/dt = -32t}}} and at t=2, we get:
{{{dh/dt = -32(2)}}} {{{dh/dt = v}}}
{{{v = -64}}}meters/second. The negative indicates a downward direction.
ii) Find the acceleration at t = 2.
Acceleration, which is the rate of change of velocity with respect to time, t, can be found by taking the second derivative of the height function ({{{h(t) = 100-16t^2}}}):
{{{d^2t/dt^2 = d(dh/dt)/dt}}}={{{d(-32t)/dt = -32}}}meters/second squared.
{{{d^2t/dt^2 = a}}}
{{{a = -32}}}meters/second squared. A constant.