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В. 5. We let y(t) be the position function. What is the position function? A. p(t) = - 16t2 - 100 t +C B. p(t) = - 16t2 - 100 t 6. What is the value of C in the position function? A. - 100 B. - 100, 000 7. What is the final position function (with the value of C substituted)? A. (t) = - 16t² – 100 t-100 B. p(t) = - 16t2 – 100 t - 100,000 C. p(t) = - 16t2 + 100 t +C D. p(t) = - 16t2 + 100 t - C C. 100 D. 100, 000 C. p(t) = - 16t2 + 100 t + 100 D. p(t) = - 16t² + 100 t + 100,000 %3D e per ho ur (66 feet per s econd

В. 5. We let y(t) be the position function. What is the position function? A. p(t) = - 16t2 - 100 t +C B. p(t) = - 16t2 - 100 t 6. What is the value of C in the position function? A. - 100 B. - 100, 000 7. What is the final position function (with the value of C substituted)? A. (t) = - 16t² – 100 t-100 B. p(t) = - 16t2 – 100 t - 100,000 C. p(t) = - 16t2 + 100 t +C D. p(t) = - 16t2 + 100 t - C C. 100 D. 100, 000 C. p(t) = - 16t2 + 100 t + 100 D. p(t) = - 16t² + 100 t + 100,000 %3D e per ho ur (66 feet per s econd

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Answer question number 5-7.
Use the word problem below to answer questions 1-7. An object's downward acceleration is given by y"(t) = -32 ft/sec2. Assume that the initial velocity is y'(0) = -100ft/s and the initial position is y(0) = 100, 000 ft. 1. Evaluate fy"(t) dt. A. -32 +C В. -32€ 2. We let V(t) the velocity function. What is V(t)? A. V(t) = -32 +C B. V(t) = -32 3. What is the value of C in the velocity function? A. - 100 B. -32 4. What is Sy'(t)dt? A. - 16t2 – 100 t + C B. - 16t2 - 100 t 5. We let y(t) be the position function. What is the position function? A. p(t) = - 16t² – 100 t + C B. p(t) = - 16t² – 100 t 6. What is the value of C in the position function? A. - 100 B. - 100, 000 7. What is the final position function (with the value of C substituted)? A. (t) = - 16t2 – 100 t - 100 B. p(t) = - 16t2 – 100 t – 100,000 8- 10. A car is travelling along a straight, level road at 45 miles per hour (66 feet per second) when the driver is forced to apply the brakes to avoid an accident. The brakes supply a constant deceleration of 22ft/sec? (feet per second, per second), how far does the car travel before coming to a complete stop? A. 99ft B. 297 ft C. -32t + C D. -32 C. V(t) = -32t + C D. V(t) = -32t C. 100 D. 32 C. - 16t2 + 100 t +C D. - 16t2 + 100 t-C В. C. p(t) = - 16t² + 100 t + C D. p(t) = - 16t² + 100 t – C %3D C. 100 D. 100, 000 C. p(t) = - 16t² + 100 t + 100 D. p(t) = - 16t² + 100 t + 100,000 C. 99 mi. D. 36 ft
Transcribed Image Text:Use the word problem below to answer questions 1-7. An object's downward acceleration is given by y"(t) = -32 ft/sec2. Assume that the initial velocity is y'(0) = -100ft/s and the initial position is y(0) = 100, 000 ft. 1. Evaluate fy"(t) dt. A. -32 +C В. -32€ 2. We let V(t) the velocity function. What is V(t)? A. V(t) = -32 +C B. V(t) = -32 3. What is the value of C in the velocity function? A. - 100 B. -32 4. What is Sy'(t)dt? A. - 16t2 – 100 t + C B. - 16t2 - 100 t 5. We let y(t) be the position function. What is the position function? A. p(t) = - 16t² – 100 t + C B. p(t) = - 16t² – 100 t 6. What is the value of C in the position function? A. - 100 B. - 100, 000 7. What is the final position function (with the value of C substituted)? A. (t) = - 16t2 – 100 t - 100 B. p(t) = - 16t2 – 100 t – 100,000 8- 10. A car is travelling along a straight, level road at 45 miles per hour (66 feet per second) when the driver is forced to apply the brakes to avoid an accident. The brakes supply a constant deceleration of 22ft/sec? (feet per second, per second), how far does the car travel before coming to a complete stop? A. 99ft B. 297 ft C. -32t + C D. -32 C. V(t) = -32t + C D. V(t) = -32t C. 100 D. 32 C. - 16t2 + 100 t +C D. - 16t2 + 100 t-C В. C. p(t) = - 16t² + 100 t + C D. p(t) = - 16t² + 100 t – C %3D C. 100 D. 100, 000 C. p(t) = - 16t² + 100 t + 100 D. p(t) = - 16t² + 100 t + 100,000 C. 99 mi. D. 36 ft
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