Questions: The speed of an object as a function of time is given by the following equation: v^2 = (8 m^2/s^4) t^2 What is the magnitude of the object's acceleration (in m / s^2 ) at t=3.8 s?

Solution Steps

The given equation for the speed of an object as a function of time is: \[ v^{2} = \left(8 \frac{m^{2}}{s^{4}}\right) t^{2} \]

To find the velocity \( v \), we take the square root of both sides of the equation: \[ v = \sqrt{8 \frac{m^{2}}{s^{4}} \cdot t^{2}} = \sqrt{8} \cdot \frac{m}{s^{2}} \cdot t \] \[ v = \sqrt{8} \cdot t \cdot \frac{m}{s^{2}} \] \[ v = 2\sqrt{2} \cdot t \cdot \frac{m}{s^{2}} \]

Acceleration \( a \) is the time derivative of velocity \( v \): \[ a = \frac{d}{dt} \left(2\sqrt{2} \cdot t \cdot \frac{m}{s^{2}}\right) \] \[ a = 2\sqrt{2} \cdot \frac{d}{dt} \left(t \cdot \frac{m}{s^{2}}\right) \] \[ a = 2\sqrt{2} \cdot \frac{m}{s^{2}} \]

Since the acceleration \( a \) is constant and does not depend on \( t \): \[ a = 2\sqrt{2} \cdot \frac{m}{s^{2}} \] \[ a = 2 \cdot 1.4142 \cdot \frac{m}{s^{2}} \] \[ a = 2.8284 \cdot \frac{m}{s^{2}} \]

Final Answer