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Chapter 2: Problem 8

Starting from rest, an object undergoes acceleration given by \(a=b t,\) where \(t\) is time and \(b\) is a constant. Can you use \(b t\) for \(a\) in Equation 2.10 to predict the object's position as a function of time? Why or why not?

Short Answer

Expert verified
No, we cannot directly use the Equation 2.10, as that equation works only when the acceleration is constant. However, In this case, acceleration is given as a function of time, thus it changes with time. We need to use integration to calculate first velocity and then position, as the acceleration function is time-dependent. The position of the object as a function of time is found to be \( b*t^3/6 \).

Step by step solution

01

Interpreting the given acceleration formula

The acceleration of an object is given as a function of time, a = bt. Here, the acceleration is not constant because it is a function of time. This means the acceleration changes with the change in time.
02

Calculating the velocity from the acceleration

Since the acceleration is the derivative of velocity with respect to time, to calculate the velocity from the acceleration, we integrate the acceleration with respect to time. The process of integration describes summation over discrete units into a continuous output. Hence, velocity at time 't', v(t) would be found by integrating a(t) with respect to time from 0 to t. \( v = \int_0^t a dt = \int_0^t bt dt = 0.5 * b*t^2 \) + C. To find C we know that at t=0, the object was at rest so v(0)=0 which implies C=0. Therefore, the velocity v = 0.5 * b * t^2.
03

Calculating the position from the velocity

Likewise, the displacement is the integral of velocity with respect to time. So, to calculate the displacement from the velocity, we integrate the velocity with respect to time. \( s = \int_0^t v dt = \int_0^t 0.5 * b * t^2 dt = b*t^3/6 \) + Ct. To find C, we know initially at t = 0 the object was at starting point, therefore C = 0. Thus displacement s = b*t^3/6.

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