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A stone is tossed into the air from ground level with an initial velocity of 31 m/s. Its height at time t is h(t) = 31t − 4.9t2 m. Compute the stone's average velocity over the time intervals [2, 2.01], [2, 2.001], [2, 2.0001], and [1.99, 2], [1.999, 2], [1.9999, 2]. (Round your answers to three decimal places.)and Estimate the instantaneous velocity v at t = 2

Question

A stone is tossed into the air from ground level with an initial velocity of 31 m/s. Its height at time t is 

h(t) = 31t − 4.9t2

 m. Compute the stone's average velocity over the time intervals 

[2, 2.01], [2, 2.001], [2, 2.0001],

 and 

[1.99, 2], [1.999, 2], [1.9999, 2].

 (Round your answers to three decimal places.)

and Estimate the instantaneous velocity v at 

t = 2
check_circleAnswer
Step 1

We use rate of change formula. 

Average velocity in time interval [2,2.01] = 11.351 m/s

h(2.01)-h(2)
2.01-2
[31(2.01)-4.9(2.012)][31(2)-4.9(22)]
0.01
42.51351-42.4
0.01
0.11351
0.01
=11.351
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Image Transcriptionclose

h(2.01)-h(2) 2.01-2 [31(2.01)-4.9(2.012)][31(2)-4.9(22)] 0.01 42.51351-42.4 0.01 0.11351 0.01 =11.351

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Step 2

At [2,2.001] average velocity is = 11.395 m/s

Using same form at [2,2.0001] average velocity is= 11.400 m/s

h(2.001)-h(2)
2.001-2
[31(2.001)-4.9(2.0012)] [31(2)-4.9(22)]
0.001
0.011395
0.001
- 11.3951
h(2.0001)-h(2)
2.0001-2
[31(2.0001)-4.9(2.00012)][31(2)-4.9(22)]
0.0001
0.001139951
0.0001
=11.39951
help_outline

Image Transcriptionclose

h(2.001)-h(2) 2.001-2 [31(2.001)-4.9(2.0012)] [31(2)-4.9(22)] 0.001 0.011395 0.001 - 11.3951 h(2.0001)-h(2) 2.0001-2 [31(2.0001)-4.9(2.00012)][31(2)-4.9(22)] 0.0001 0.001139951 0.0001 =11.39951

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Step 3

Average velocity in [1.99...

h(2)-h(1.99)
2-1.99
[31(2)-4.9(22)]-[31(1.99)-4.9(1.99)]
0.01
0.11449
0.01
11.449
help_outline

Image Transcriptionclose

h(2)-h(1.99) 2-1.99 [31(2)-4.9(22)]-[31(1.99)-4.9(1.99)] 0.01 0.11449 0.01 11.449

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Tagged in

Math

Calculus

Derivative

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