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Chapter 3: Problem 23

An object's velocity is \(\vec{v}=c t^{3} \hat{\imath}+d \hat{\jmath},\) where \(t\) is time and \(c\) and \(d\) are positive constants with appropriate units. What's the direction of the object's acceleration?

Short Answer

Expert verified
The direction of the object's acceleration is along the positive x direction.

Step by step solution

01

Understand the Problem and Identify Knowns

The object's velocity is given by the equation \(\vec{v}=c t^{3} \hat{\imath}+d \hat{\jmath}\), where \(t\) is time and \(c\) and \(d\) are positive constants. We need to find the direction of the object's acceleration, which is the derivative of the velocity with respect to time.
02

Calculate the Derivative of the Velocity

Taking the derivative of the velocity with respect to time, we get the acceleration \(\vec{a} = \frac{d\vec{v}}{dt}\). For the \(i\) component, by applying the Power Rule for differentiation, we have \(\frac{d}{dt}(c t^{3}) = 3ct^{2}\). For the \(j\) component, the derivative of a constant is zero. Therefore, the acceleration is \(\vec{a} = 3ct^{2} \hat{\imath}\).
03

Determine the Direction of Acceleration

The direction of the acceleration vector \(\vec{a}\) is along the x-axis (i.e., the direction of the unit vector \(\hat{\imath}\)). Since \(3ct^{2}\) is always positive for positive \(c\) and \(t\), the acceleration vector points in the positive x direction.

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