Question

A small object moves along the x-axis with acceleration ax(t) = -(0.0320 m/s^3)(15.0 s - t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.20 m/s. Part A What is the x-coordinate of the object when t = 10.0 s? Express your answer with the appropriate units.

          A small object moves along the x-axis with acceleration ax(t) = -(0.0320 m/s^3)(15.0 s - t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.20 m/s.

Part A
What is the x-coordinate of the object when t = 10.0 s?
Express your answer with the appropriate units.

A small object moves along the x-axis with acceleration ax(t) = -(0.0320 m/s^3)(15.0 s - t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.20 m/s.

Part A
What is the x-coordinate of the object when t = 10.0 s?
Express your answer with the appropriate units.

Added by Alexander A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Mkhitar Hobosyan

09/03/2023

Step 1

Given acceleration function: $a_x(t) = 0.0320(15.0t)$ Integrating the acceleration function to get the velocity function: $v(t) = \int a_x(t) dt$ $v(t) = \int 0.0320(15.0t) dt$ $v(t) = 0.0320 \times \frac{15.0}{2}t^2$ $v(t) = 0.240t^2$ Show more…

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A small object moves along the x-axis with acceleration ax(t) = -(0.0320 m/s^3)(15.0 s - t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.20 m/s. Part A What is the x-coordinate of the object when t = 10.0 s? Express your answer with the appropriate units.