Can someone help me with v(t)? If a stone is thrown up at 13 meters per second from a height of 190 meters above the surface of the moon, its height in meters after t seconds is given by s = 190 + 13t - 0.8t². What is its acceleration?Step 1Recall 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 ofchange, acceleration is the derivative of velocity a(t) = v'(t).So, to determine acceleration, first find velocity.Given that s(t) = 190 + 13t - 0.8t², the velocity v(t) is found by taking the first ✔1.6 t, we have v(t) = 1.6Since s'(t) = 13 -1.6✔XStep 2Next, acceleration, a(t), is found by taking the first derivative of velocity, v'(t),So, since v(t) = 13- 1.6t we have v'(t) =Thus, acceleration of the stone on the moon is a(t) = -1.613 1.6t✔ m/s².first derivative of position, s'(t).

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If a stone is thrown up at 13 meters per second from a height of 190 meters above the surface of the moon, its height in meters after t seconds is given by s = 190 + 13t - 0.8+2. 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 change, acceleration is the derivative f velocity a(t)= v'(t). So, to determine acceleration, first find velocity. Given that s(t) = 190 + 13t - 0.8t², the velocity v(t) is found by taking the first ✔ 1.6 t, we have v(t) = 1.6 Since s'(t) = 13 -1.6 x Step 2 Next, acceleration, a(t), is found by taking the first derivative of velocity, v'(t). So, since v(t) = 13 - 1.6t we have v'(t) = Thus, acceleration of the stone on the moon is a(t) = -1.6 131.6t ✔ m/s². first derivative of position, s'(t). changing is the acceleration. Because the derivative measures the rate of

If a stone is thrown up at 13 meters per second from a height of 190 meters above the surface of the moon, its height in meters after t seconds is given by s = 190 + 13t - 0.8+2. 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 change, acceleration is the derivative f velocity a(t)= v'(t). So, to determine acceleration, first find velocity. Given that s(t) = 190 + 13t - 0.8t², the velocity v(t) is found by taking the first ✔ 1.6 t, we have v(t) = 1.6 Since s'(t) = 13 -1.6 x Step 2 Next, acceleration, a(t), is found by taking the first derivative of velocity, v'(t). So, since v(t) = 13 - 1.6t we have v'(t) = Thus, acceleration of the stone on the moon is a(t) = -1.6 131.6t ✔ m/s². first derivative of position, s'(t). changing is the acceleration. Because the derivative measures the rate of

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter3: The Derivative

Section3.CR: Chapter 3 Review

Problem 8CR

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