Free-Falling Object question, halfway done

[VERON!CA]

Use the position function s(t) = -16t^t + 1000, which gives the height (in feet) of an object that has fallen for t seconds from a height of 1000 feet. The velocity at time t = a seconds is:

. . .. .lim. . .[ s(a) - s(t) ] / (a - t)
. . .t --> a

If a construction worker drops a wrench from a height of 1000 feet, how fast will the wrench be falling after 5 seconds?

I think I have half of the problem done, but I'm not sure what to do afterwards.

This is my work so far:

. . .s(t) = -16(5)^2 + 1000

. . .s(t) = 600

 

[stapel]

On what basis have you fixed s(t) at a value of 600 (in your last line)? By definition, the position function should be variable, since the object is moving.

A good start might be to follow the instructions: Find the velocity function by finding the limit (given) that provides that function. Plug s(t) into the provided expression, and take the listed limit. Then use the result (which is the velocity function) to find the velocity at t = 5.

Eliz.

 

[TchrWill]

Use the position function s(t) = -16t^t + 1000, which gives the height (in feet) of an object that has fallen for t seconds from a height of 1000 feet. The velocity at time t = a seconds is:

. . .. .lim. . .[ s(a) - s(t) ] / (a - t)
. . .t --> a

If a construction worker drops a wrench from a height of 1000 feet, how fast will the wrench be falling after 5 seconds?

I think I have half of the problem done, but I'm not sure what to do afterwards.

The distance actually fallen in t seconds is defined by d = Vot + 16t^2 or 16t^2 with an initial velocity of Vo = 0.

The position of an object dropped from a height of 1000 ft. is given by s = 1000 - Vot - 16t^2, or 1000 - 16t^2 for Vo = 0, s being the distance from the ground.

The equation s(t) = Vot - 16t^2 is the height reached when propelled upward at a velocity of Vo.

An object dropped from a height if 1000 ft. will have a velocity of Vf = Vo + gt = Vo + 32t = 0 + 32(5)= 160 ft/sec.

The distance fallen in this 5 seconds is s = Vot + gt^2/2 = Vot + 16t^2 = 0 + 16(5^2) = 400 feet, making the position, or distance, from the ground 600 feet.

 

[skeeter]

Use the position function s(t) = -16t^t + 1000, which gives the height (in feet) of an object that has fallen for t seconds from a height of 1000 feet. The velocity at time t = a seconds is:

. . .. .lim. . .[ s(a) - s(t) ] / (a - t)
. . .t --> a

If a construction worker drops a wrench from a height of 1000 feet, how fast will the wrench be falling after 5 seconds?

I think I have half of the problem done, but I'm not sure what to do afterwards.

This is my work so far:

. . .s(t) = -16(5)^2 + 1000

it should be ...

s(5) = -16(5)² + 1000 = 600

now use the limit to evaluate the instantaneous velocity at t = 5 sec ...

lim{t->5} [s(5) - s(t)]/(5 - t) =

lim{t->5} [600 - (-16t² + 1000)]/(5 - t) =

lim{t->5} [16t² - 400]/(5 - t) =

16*lim{t->5} [t² - 25]/(5 - t) =

16*lim{t->5} [(t + 5)(t - 5)]/(5 - t) =

16*lim{t->5} -(t + 5) = -160 ft/sec