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The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of the pebble above the water surface after t seconds is given by y = 235 − 16t^2.

Required:
a. Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time.

1. 0.1 sec:________
2. 0.05 sec:_______
3. 0.01 sec:_______

b. Estimate the instantaneous velocity (in ft/s) of the pebble after 3 seconds. ft/s.

Respuesta :

onyebuchinnaji

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

[tex]y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}[/tex]

(1)

[tex]y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s[/tex]

(2)

[tex]y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s[/tex]

(3)

[tex]y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s[/tex]

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

(a)The average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

What is the average velocity?

The total displacement traveled by an object divided by the total time taken is the average velocity.

Average velocity is given by;

[tex]\rm y(t_2,t_1)=\frac{y(t_2)-y(t_1)}{t_2-t_1} \\\\[/tex]

The average velocity for case 1;

[tex]\rm y(2.1,2)=\frac{(235-16\times 2\times 1^2)-(235-16 \times 2^2)}{2.1-2} \\\\ \rm y(2.1,2)=\-65.6 \ m/sec[/tex]

The average velocity for case 2;

[tex]\rm y(2.015,2)=\frac{(235-16\times 2.05^2)-(235-16 \times 2^2)}{2.05-2} \\\\ \rm y(2.1,2)=\-64.8 \ m/sec[/tex]

The average velocity for case 3;

[tex]\rm y(2.01,2)=\frac{(235-16\times 2.01^2)-(235-16 \times 2^2)}{2.01-2} \\\\ \rm y(2.1,2)=\-64.16 \ m/sec[/tex]

Hence the average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

(b) The instantaneous velocity will be -96 m/s.

The given equation in the problem is;

[tex]\rm y = 235 - 16t^2[/tex]

The instantaneous velocity is given as;

[tex]v = \frac{dy}{dt} \\\\ \frac{dy}{dt} = -32t\\\\ t = 3 s\\\\ v = -32\times 3\\\\ v = -96 m/s[/tex]

Hence the instantaneous velocity will be -96 m/s.

To learn more about the average velocity refer to the link;

https://brainly.com/question/862972

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