Question

If an arrow is shot upward on Mars with a speed of 54 m/s, its height in meters t seconds later is given by

Answer 1

1) Average speed in the different intervals:

48.42 m/s

49.36 m/s

50.10 m/s

50.0 m/s

50.0 m/s

2) Instantaneous speed at t = 1 s is 50.28 m/s

Step-by-step explanation:

1)

The position of the arrow at time t is given by the equation

$y(t)=54t-1.86t^2$

The average speed can be found by dividing the distance covered by the time taken:

$v=\frac{\Delta y}{\Delta t}$

(i) For this interval [1,2]:

$y(1)=54(1)-1.86(1)^2=52.14 m\\y(2)=54(2)-1.86(2)^2=100.56 m$

So $\Delta y = y(2)-y(1)=100.56-52.14=48.42 m$

Time interval is $\Delta t = 2-1 = 1 s$

So the average speed is

$v=\frac{48.42}{1}=48.42 m/s$

(II) For this interval [1,1.5]:

$y(1)=54(1)-1.86(1)^2=52.14 m\\y(1.5)=54(1.5)-1.86(1.5)^2=76.82 m$

So $\Delta y = y(1.5)-y(1)=76.82-52.14=24.68 m$

Time interval is $\Delta t = 1.5-1 = 0.5 s$

So the average speed is

$v=\frac{24.68}{0.5}=49.36 m/s$

(iii) For this interval [1,1.1]:

$y(1)=54(1)-1.86(1)^2=52.14 m\\y(1.5)=54(1.1)-1.86(1.1)^2=57.15 m$

So $\Delta y = y(1.1)-y(1)=57.15-52.14=5.01 m$

Time interval is $\Delta t = 1.1-1 = 0.1 s$

So the average speed is

$v=\frac{5.01}{0.1}=50.10 m/s$

(iv) For this interval [1,1.01]:

$y(1)=54(1)-1.86(1)^2=52.14 m\\y(1.5)=54(1.01)-1.86(1.01)^2=52.64 m$

So $\Delta y = y(1.01)-y(1)=52.64-52.14=0.50 m$

Time interval is $\Delta t = 1.01-1 = 0.01 s$

So the average speed is

$v=\frac{0.50}{0.01}=50.0 m/s$

(v) For this interval [1,1.001]:

$y(1)=54(1)-1.86(1)^2=52.14 m\\y(1.5)=54(1.001)-1.86(1.001)^2=52.19 m$

So $\Delta y = y(1.001)-y(1)=52.19-52.14=0.05 m$

Time interval is $\Delta t = 1.001-1 = 0.001 s$

So the average speed is

$v=\frac{0.05}{0.001}=50.0 m/s$

2)

In this part we want to estimate the instantaneous speed of the arrow. The instantaneous speed can be found by calculating the derivative of the position, therefore:

$y(t)=54t-1.86t^2$

So the derivative is

$v(t)=y'(t)=54-2\cdot 1.86t=54-3.72t$

And by substituting t = 1 s, we find the instantaneou speed at that time:

$v(1)=54-3.72(1)=50.28 m/s$

Learn more about speed:

brainly.com/question/8893949

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