A particle moves along a straight line and passes through a fixed point O . It’s acceleration, a [math]ms^{-2}[/math] , is given by [math]a=2t−6[/math] , where t is the time, in seconds, after passing through O . The initial velocity of the particle is 8 [math]ms^{-1}[/math]. Find [br][Assume motion to the right means positive velocity][br][br]a) the minimum velocity of the particle in ms[sup]-1[/sup]
The acceleration is the derivative of velocity with respect to time:[br][math]a\left(t\right)=\frac{dv}{dt}[/math][br][br]So, to get the velocity function, we integrate the acceleration function with respect to time:[br][math]v\left(t\right)=\int\left(2t-6\right)dt[/math][br][math]v\left(t\right)=t^2-6t+C[/math][br]Here, C is the constant of integration, which we can determine using the initial velocity condition. We are told that the initial velocity at t=0 is v(0) = 8ms[sup]-1[br][br][math]v\left(0\right)-0^2-6\left(0\right)+C=8[/math][br][math]v\left(t\right)=t^2-6t+8[/math][br][br][size=100]The velocity will be minimum when the derivative of the velocity function (i.e., the acceleration) is zero. This is because the particle changes direction when the velocity reaches its minimum.[br]We already have the acceleration function:[/size][br][/sup]a(t) = 2t - 6[br]Set a(t) = 0 to find the time when the velocity is minimum:[br]2t - 6 = 0[br]t=3s[br][br]Now that we know the time t=3 seconds, substitute this value into the velocity equation to find the minimum velocity:[br][math]v\left(3\right)=\left(3\right)^2-6\left(3\right)+8=-1ms^{-1}[/math][br][br]Final Answer:[br][br]The minimum velocity of the particle is -1ms[sup]-1[/sup]