Tutorial Exercise If a stone is thrown up at 17 meters per second from a height of 170 meters above the surface of the moon, its height in meters after t seconds is given by s = 170 + 17t - 0.8t. What is its acceleration? Step 1 Recall that if s(t) represents the position at time t of any object moving in a straight line, then its velocity is given by the derivative v(t) = s'(t). Now, suppose that object is a car. Since a car rarely drives at a constant speed, the velocity itself may be changing. The rate at which the velocity is changing is the acceleration. Because the derivative measures the rate of change, acceleration is the derivative of velocity a(t) = v'(t). So, to determine acceleration, first find velocity. Given that s(t) = 170 + 17t – 0.8t, the velocity v(t) is found by taking the first position, s'(t). derivative of Since s'(t) = 17 - t, we have v(t) =

Tutorial Exercise If a stone is thrown up at 17 meters per second from a height of 170 meters above the surface of the moon, its height in meters after t seconds is given by s = 170 + 17t - 0.8t. What is its acceleration? Step 1 Recall that if s(t) represents the position at time t of any object moving in a straight line, then its velocity is given by the derivative v(t) = s'(t). Now, suppose that object is a car. Since a car rarely drives at a constant speed, the velocity itself may be changing. The rate at which the velocity is changing is the acceleration. Because the derivative measures the rate of change, acceleration is the derivative of velocity a(t) = v'(t). So, to determine acceleration, first find velocity. Given that s(t) = 170 + 17t – 0.8t, the velocity v(t) is found by taking the first position, s'(t). derivative of Since s'(t) = 17 - t, we have v(t) =

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:HOUGHTON MIFFLIN HARCOURT

Chapter8: Graphing Quadratic Functions

Section: Chapter Questions

Problem 30CR

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