Question

1. the velocity of a particle moving along the x - axis is given by ( v(t)=sin(2t) ) at time ( t ). if the particle is at ( x = 4 ) when ( t = 0 ), what is the position of the particle when ( t=\frac{pi}{2} )?

(a) 2
(b) 3
(c) 4
(d) 5
(e) 6

Explanation:

Step1: Find the position function

We know that \(v(t)=\sin(2t)\), and \(x(t)=\int v(t)dt\). Using the integral formula \(\int\sin(at)dt =-\frac{1}{a}\cos(at)+C\) (here \(a = 2\)), we have \(x(t)=-\frac{1}{2}\cos(2t)+C\).

Step2: Determine the constant \(C\)

Since \(x(0) = 4\), substitute \(t = 0\) into \(x(t)\): \(4=-\frac{1}{2}\cos(0)+C\). Since \(\cos(0)=1\), then \(4=-\frac{1}{2}+C\), so \(C = 4+\frac{1}{2}=\frac{9}{2}\).

Step3: Calculate \(x(\frac{\pi}{2})\)

Substitute \(t=\frac{\pi}{2}\) into \(x(t)=-\frac{1}{2}\cos(2t)+\frac{9}{2}\). We know that \(\cos(2\times\frac{\pi}{2})=\cos(\pi)=- 1\). Then \(x(\frac{\pi}{2})=-\frac{1}{2}\times(-1)+\frac{9}{2}=\frac{1 + 9}{2}=5\).

Answer:

D. 5