Forgot how to solve for velocity in mathematics -_-

A-R-Q · **Joined:** Sat, 20 Nov 2010 21:17:02 UTC **Posts:** 432

The height in metres, of an object that has fallen from a height of 180m is given by the position function s(t) = -5t^2 + 180, when t greater than or equal to 0 and t is in seconds.

c) At what velocity will the object hit the ground?

How would you solve having previous knowledge of average rate of change, instantaneous rate of change, limits? I know how to get it using the kinematics equations, but I am having trouble doing this through pure math.

I know that d will equal 0 when this happens, and t I don't know. I solve for t with d being 0 and I get (t-6)(t+6), so t= 6s.

Shadow · **Posted:** Sat, 16 Apr 2011 19:36:46 UTC

A-R-Q wrote:

The height in metres, of an object that has fallen from a height of 180m is given by the position function s(t) = -5t^2 + 180, when t greater than or equal to 0 and t is in seconds.

c) At what velocity will the object hit the ground?

How would you solve having previous knowledge of average rate of change, instantaneous rate of change, limits? I know how to get it using the kinematics equations, but I am having trouble doing this through pure math.

I know that d will equal 0 when this happens, and t I don't know. I solve for t with d being 0 and I get (t-6)(t+6), so t= 6s.

the veolocity at a time, t, is $$\lim_{t\to 6} {f(t)-f(6)\over t-6}$

A-R-Q · **Joined:** Sat, 20 Nov 2010 21:17:02 UTC **Posts:** 432

Shadow wrote:

A-R-Q wrote:

The height in metres, of an object that has fallen from a height of 180m is given by the position function s(t) = -5t^2 + 180, when t greater than or equal to 0 and t is in seconds.

c) At what velocity will the object hit the ground?

How would you solve having previous knowledge of average rate of change, instantaneous rate of change, limits? I know how to get it using the kinematics equations, but I am having trouble doing this through pure math.

I know that d will equal 0 when this happens, and t I don't know. I solve for t with d being 0 and I get (t-6)(t+6), so t= 6s.

the veolocity at a time, t, is $$\lim_{t\to 6} {f(t)-f(6)\over t-6}$

Ok thank you, but I thought velocity is like this:

velocity = displacement / time
In this case if the object hit the ground, wouldn't the displacement be 0, and thus the numerator be 0? I am thinking this because when it hits the ground, shouldn't the given function = 0?

Shadow · **Posted:** Sat, 16 Apr 2011 19:51:59 UTC

A-R-Q wrote:

Shadow wrote:

A-R-Q wrote:

The height in metres, of an object that has fallen from a height of 180m is given by the position function s(t) = -5t^2 + 180, when t greater than or equal to 0 and t is in seconds.

c) At what velocity will the object hit the ground?

How would you solve having previous knowledge of average rate of change, instantaneous rate of change, limits? I know how to get it using the kinematics equations, but I am having trouble doing this through pure math.

I know that d will equal 0 when this happens, and t I don't know. I solve for t with d being 0 and I get (t-6)(t+6), so t= 6s.

the veolocity at a time, t, is $$\lim_{t\to 6} {f(t)-f(6)\over t-6}$

Ok thank you, but I thought velocity is like this:

velocity = displacement / time
In this case if the object hit the ground, wouldn't the displacement be 0, and thus the numerator be 0? I am thinking this because when it hits the ground, shouldn't the given function = 0?

it's a limit, you're not actually dividing by 0. You said you had knowledge of limits.

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Forgot how to solve for velocity in mathematics -_-

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