Question

If an arrow is shot upward on Mars with a speed of 58 m/s, its height in meters t seconds later is given by y = 58t − 1.86t2. (Round your answers to two decimal places.) (a) Find the average speed over the given time intervals.
i. t= [1, 2]
ii. t=[1, 1.5]
iii. [1, 1.1]
iv. [1, 1.01]

Answer 1

Solution-

The speed 58 m/s with which the ball was shot is the extra data given, as it is already included in the equation as the co-efficient of t .

The height of the arrow (in meters) after t sec is given by

$y = 58t-1.86t^2$

$Average \ speed=\frac{Distance}{Time}$

i.

Height of ball after t=1 sec,

$y(1) = 58(1)-1.86(1)^2=56.14$ (putting t=1 in the height equation)

Height of ball after t=2 sec,

$y(2)= 58(2)-1.86(2)^2=108.64$ (putting t=2 in the height equation)

∴ Distance traveled in between

$y(2)-y(1)=108.56-56.14=52.42 \ m$

And Time taken

$2-1=1 \ sec$

$\therefore Avg.speed=\frac{52.42}{1} =52.42 \ m/s$

ii.

Height of ball after t=1 sec,

$y(1) = 58(1)-1.86(1)^2=56.14$ (putting t=1 in the height equation)

Height of ball after t=1.5 sec,

$y(1.5)= 58(1.5)-1.86(1.5)^2=82.815$ (putting t=1.5 in the height equation)

∴ Distance traveled in between

$y(1.5)-y(1)=82.81-56.14=23.67\ m$

And Time taken

$1.5-1=0.5 \ sec$

$\therefore Avg.speed=\frac{26.67}{0.5} =53.35 \ m/s$

iii.

Height of ball after t=1 sec,

$y(1) = 58(1)-1.86(1)^2=56.14$ (putting t=1 in the height equation)

Height of ball after t=1.1 sec,

$y(1.1)= 58(1.1)-1.86(1.1)^2=61.5494$ (putting t=1.1 in the height equation)

∴ Distance traveled in between

$y(1.1)-y(1)=61.5494-56.14=5.4094\ m$

And Time taken

$1.1-1=0.1 \ sec$

$\therefore Avg.speed=\frac{5.4094}{0.1} =54.094 \approx 54.09 \ m/s$

iv.

Height of ball after t=1 sec,

$y(1) = 58(1)-1.86(1)^2=56.14$ (putting t=1 in the height equation)

Height of ball after t=1.01 sec,

$y(1.01)= 58(1.01)-1.86(1.01)^2=56.6826$ (putting t=1.01 in the height equation)

∴ Distance traveled in between

$y(1.01)-y(1)=56.6826-56.14=.5426\ m$

And Time taken

$1.01-1=0.01 \ sec$

$\therefore Avg.speed=\frac{.5426}{0.01} =54.26 \ m/s$

Answer 2

The distance is given by
y = 58t - 1.86t²

The velocity is the derivative of y with respect to t. It is
v = 58 - 3.72t

In an interval [t₁, t₂], the average velocity is
$\bar{v} = \frac{1}{t_{2}-t_{1}}[58t-1.86t^{2}]_{t1}^{t2}$

Evaluate the results (Use v to denote the average velocity).
i. t = [1,2]
   v = (108.56 - 56.14)/1 = 52.42
   Answer: 52.42 m/s

ii. t=[1,1.5]
   v = (82.815 - 56.14)/0.5 = 53.35 
   Answer: 53.35 m/s

iii. t=[1,1.1]
    v = (61.5494 - 56.14)/0.1 = 54.094
    Answer: 54.094 m/s

iv. t=[1,1.01]
   v = (56.6826 - 56.14)/0.01 = 54.26
   Answer: 54.26 m/s

Note:
As the time interval approaches zero, we obtain the derivative of y at t=1, which is 58-3.72 = 54.28 m/s