A ball's velocity is given by $v\left(t\right)=\frac{1}{2}{t}^3-2t$. Calculate the ball's displacement from $t=1.5s$ to $t=3s$.

• A. $8.81m$
• B. $2.74m$
• C. $35.0m$
• D. $-2.74m$

Answer: (B)

Verified step by step guidance

1. Identify that displacement is the integral of velocity over the given time interval. The displacement \(\Delta x\) from time \(t_1\) to \(t_2\) is given by \(\Delta x = \int_{t_1}^{t_2} v(t) \, dt\).
2. Write down the given velocity function: \(v(t) = \frac{1}{2} t^3 - 2t\).
3. Set up the definite integral for displacement from \(t = 1.5\) s to \(t = 3\) s: \(\Delta x = \int_{1.5}^{3} \left( \frac{1}{2} t^3 - 2t \right) dt\).
4. Integrate the velocity function term-by-term: the integral of \(\frac{1}{2} t^3\) is \(\frac{1}{2} \cdot \frac{t^4}{4} = \frac{t^4}{8}\), and the integral of \(-2t\) is \(-2 \cdot \frac{t^2}{2} = -t^2\).
5. Evaluate the resulting expression \(\left[ \frac{t^4}{8} - t^2 \right]\) at the upper limit \(t=3\) and the lower limit \(t=1.5\), then subtract to find the displacement: \(\Delta x = \left( \frac{3^4}{8} - 3^2 \right) - \left( \frac{(1.5)^4}{8} - (1.5)^2 \right)\).