A particle with velocity \( v(t)=3 \sqrt{t} \), where \( t \) is time in seconds, moves in a straight line. How far does the particle move from \( t=1 \) to \( t=4 \) seconds?

To find the distance the particle moves from \( t=1 \) to \( t=4 \) seconds, we need to compute the integral of the velocity function over the given time interval. The velocity function is given by: \[ v(t) = 3 \sqrt{t} \] The distance \( d \) traveled from \( t=1 \) to \( t=4 \) is: \[ d = \int_{1}^{4} v(t) \, dt = \int_{1}^{4} 3 \sqrt{t} \, dt \] Calculating the integral, we have: \[ d = 3 \int_{1}^{4} t^{1/2} \, dt \] The integral of \( t^{1/2} \) is: \[ \int t^{1/2} \, dt = \frac{2}{3} t^{3/2} \] Now substituting the limits: \[ d = 3 \left[ \frac{2}{3} t^{3/2} \right]_{1}^{4} = 3 \left( \frac{2}{3} (4^{3/2} - 1^{3/2}) \right) \] Calculating \( 4^{3/2} = 8 \) and \( 1^{3/2} = 1 \): \[ d = 3 \left( \frac{2}{3} (8 - 1) \right) = 3 \left( \frac{2}{3} \times 7 \right) = 3 \times \frac{14}{3} = 14 \] Thus, the particle moves a distance of \( 14 \) units from \( t=1 \) to \( t=4 \) seconds.