QuestionA ball is thrown with a velocity of $41 \mathrm{ft/s}$ . Height after $t$ seconds is $y=41t-22t^2$ . Find average velocity from $t=2$ for 0.01, 0.005, and 0.002 seconds. What is the instantaneous velocity at $t=2$ ?

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the ball is $41 \, \mathrm{ft/s}$ . The height of the ball after $t$ seconds is given by $y=41t-22t^{}$
3. We need to find the average velocity for the time periods beginning when $t=$ for the following time intervals0.01 seconds,0.005 seconds,0.002 seconds

STEP 2

The average velocity for a time period can be calculated by taking the difference in the position (height in this case) at the end and start of the time period, divided by the duration of the time period. $\text{Average velocity} = \frac{\text{Change in position}}{\text{Change in time}}$

STEP 3

We can express the change in position as $y(t+\Delta t) - y(t)$ , where $y(t)$ is the position at time $t$ , $y(t+\Delta t)$ is the position at time $t+\Delta t$ , and $\Delta t$ is the change in time.

STEP 4

Substitute the given function $y=41t-22t^{2}$ into the formula for the change in position.
$\text{Change in position} = y(t+\Delta t) - y(t) = (41(t+\Delta t)-22(t+\Delta t)^{2}) - (41t-22t^{2})$

STEP 5

Calculate the average velocity for the time period beginning when $t=2$ for $\Delta t =0.01$ seconds.
$\text{Average velocity}_{0.01} = \frac{41(2+0.01)-22(2+0.01)^{2} - (41(2)-22(2)^{2})}{0.01}$

STEP 6

Calculate the average velocity for the time period beginning when $t=2$ for $\Delta t =0.005$ seconds.
$\text{Average velocity}_{0.005} = \frac{41(2+0.005)-22(2+0.005)^{2} - (41(2)-22(2)^{2})}{0.005}$

STEP 7

Calculate the average velocity for the time period beginning when $t=2$ for $\Delta t =0.002$ seconds.
$\text{Average velocity}_{0.002} = \frac{41(2+0.002)-22(2+0.002)^{2} - (41(2)-22(2)^{2})}{0.002}$

STEP 8

The instantaneous velocity when $t=2$ is the limit of the average velocity as $\Delta t$ approaches0. $\text{Instantaneous velocity} = \lim_{\Delta t \to0} \frac{41(2+\Delta t)-22(2+\Delta t)^{2} - (41(2)-22(2)^{2})}{\Delta t}$

STEP 9

This limit is the derivative of the position function $y(t)$ at $t=2$ . The derivative of $y(t)$ is $y'(t) =41 -44t$ .

STEP 10

Substitute $t=2$ into the derivative to find the instantaneous velocity.
$\text{Instantaneous velocity} =41 -44(2)$

STEP 11

Calculate the instantaneous velocity.
$\text{Instantaneous velocity} =41 -44() = -47 \, \mathrm{ft/s}$ The instantaneous velocity when $t=$ is $-47 \, \mathrm{ft/s}$ .

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QuestionA ball is thrown with a velocity of $41 \mathrm{ft/s}$ . Height after $t$ seconds is $y=41t-22t^2$ . Find average velocity from $t=2$ for 0.01, 0.005, and 0.002 seconds. What is the instantaneous velocity at $t=2$ ?

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the ball is $41 \, \mathrm{ft/s}$ . The height of the ball after $t$ seconds is given by $y=41t-22t^{}$
3. We need to find the average velocity for the time periods beginning when $t=$ for the following time intervals0.01 seconds,0.005 seconds,0.002 seconds

STEP 2

The average velocity for a time period can be calculated by taking the difference in the position (height in this case) at the end and start of the time period, divided by the duration of the time period. $\text{Average velocity} = \frac{\text{Change in position}}{\text{Change in time}}$

STEP 3

We can express the change in position as $y(t+\Delta t) - y(t)$ , where $y(t)$ is the position at time $t$ , $y(t+\Delta t)$ is the position at time $t+\Delta t$ , and $\Delta t$ is the change in time.

STEP 4

Substitute the given function $y=41t-22t^{2}$ into the formula for the change in position.
$\text{Change in position} = y(t+\Delta t) - y(t) = (41(t+\Delta t)-22(t+\Delta t)^{2}) - (41t-22t^{2})$

STEP 5

Calculate the average velocity for the time period beginning when $t=2$ for $\Delta t =0.01$ seconds.
$\text{Average velocity}_{0.01} = \frac{41(2+0.01)-22(2+0.01)^{2} - (41(2)-22(2)^{2})}{0.01}$

STEP 6

Calculate the average velocity for the time period beginning when $t=2$ for $\Delta t =0.005$ seconds.
$\text{Average velocity}_{0.005} = \frac{41(2+0.005)-22(2+0.005)^{2} - (41(2)-22(2)^{2})}{0.005}$

STEP 7

Calculate the average velocity for the time period beginning when $t=2$ for $\Delta t =0.002$ seconds.
$\text{Average velocity}_{0.002} = \frac{41(2+0.002)-22(2+0.002)^{2} - (41(2)-22(2)^{2})}{0.002}$

STEP 8

The instantaneous velocity when $t=2$ is the limit of the average velocity as $\Delta t$ approaches0. $\text{Instantaneous velocity} = \lim_{\Delta t \to0} \frac{41(2+\Delta t)-22(2+\Delta t)^{2} - (41(2)-22(2)^{2})}{\Delta t}$

STEP 9

This limit is the derivative of the position function $y(t)$ at $t=2$ . The derivative of $y(t)$ is $y'(t) =41 -44t$ .

STEP 10

Substitute $t=2$ into the derivative to find the instantaneous velocity.
$\text{Instantaneous velocity} =41 -44(2)$

STEP 11

Calculate the instantaneous velocity.
$\text{Instantaneous velocity} =41 -44() = -47 \, \mathrm{ft/s}$ The instantaneous velocity when $t=$ is $-47 \, \mathrm{ft/s}$ .

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QuestionA ball is thrown with a velocity of 41ft/s41 \mathrm{ft/s}41ft/s. Height after ttt seconds is y=41t−22t2y=41t-22t^2y=41t−22t2. Find average velocity from t=2t=2t=2 for 0.01, 0.005, and 0.002 seconds. What is the instantaneous velocity at t=2t=2t=2?

Studdy Solution

STEP 1

STEP 2

STEP 3

We can express the change in position as y(t+Δt)−y(t)y(t+\Delta t) - y(t)y(t+Δt)−y(t), where y(t)y(t)y(t) is the position at time ttt, y(t+Δt)y(t+\Delta t)y(t+Δt) is the position at time t+Δtt+\Delta tt+Δt, and Δt\Delta tΔt is the change in time.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

This limit is the derivative of the position function y(t)y(t)y(t) at t=2t=2t=2. The derivative of y(t)y(t)y(t) is y′(t)=41−44ty'(t) =41 -44ty′(t)=41−44t.

STEP 10

Substitute t=2t=2t=2 into the derivative to find the instantaneous velocity.Instantaneous velocity=41−44(2)\text{Instantaneous velocity} =41 -44(2)Instantaneous velocity=41−44(2)

STEP 11

Calculate the instantaneous velocity.Instantaneous velocity=41−44()=−47 ft/s\text{Instantaneous velocity} =41 -44() = -47 \, \mathrm{ft/s}Instantaneous velocity=41−44()=−47ft/sThe instantaneous velocity when t=t=t= is −47 ft/s-47 \, \mathrm{ft/s}−47ft/s.

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QuestionA ball is thrown with a velocity of $41 \mathrm{ft/s}$ . Height after $t$ seconds is $y=41t-22t^2$ . Find average velocity from $t=2$ for 0.01, 0.005, and 0.002 seconds. What is the instantaneous velocity at $t=2$ ?

We can express the change in position as $y(t+\Delta t) - y(t)$ , where $y(t)$ is the position at time $t$ , $y(t+\Delta t)$ is the position at time $t+\Delta t$ , and $\Delta t$ is the change in time.

This limit is the derivative of the position function $y(t)$ at $t=2$ . The derivative of $y(t)$ is $y'(t) =41 -44t$ .

Substitute $t=2$ into the derivative to find the instantaneous velocity.
$\text{Instantaneous velocity} =41 -44(2)$

Calculate the instantaneous velocity.
$\text{Instantaneous velocity} =41 -44() = -47 \, \mathrm{ft/s}$ The instantaneous velocity when $t=$ is $-47 \, \mathrm{ft/s}$ .