A particle moves along a straight line with velocity given by v = (6 – 3t) where v is in m/s and t in seconds. Determine when the particle returns to its starting point.
Displacement is \(s = \int v \, dt = 6t - 1.5t^2\). To return to the starting point, \(s = 0 \implies 6t - 1.5t^2 = 0 \implies t = 4\text{ s}\).
A particle begins to move along straight line where the acceleration \( (a) \) of the particle varies with displacement \( (x) \) according to relation, \( a = 5x \), then velocity of the particle varies with displacement as
Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).
A particle moves along a straight line with velocity given by \( v = (6 – 3t) \) where \( v \) is in \( \text{m/s} \) and \( t \) in seconds. Determine when the particle returns to its starting point.
Displacement is \( S = \int v \, dt = \int_0^t (6 - 3t) \, dt = 6t - 1.5t^2 \). Returning to the starting point means \( S = 0 ⇒ 6t - 1.5t^2 = 0 ⇒ t = 4\text{ s} \).