[size=50][size=100]A particle moves along a straight line and passes through a fixed point with an initial velocity of 4 [math]ms[/math]. Its acceleration, a [math]ms^{-2}[/math],t seconds after passing through O is given by a = 4 - 2t. [br][br](a)calculate[br] (i) the instantaneous velocity, in me, of the particle when t = 7[br] (ii) the maximum velocity in ms–¹, of the particle.[br][br](b) Find the possible values of t, in seconds, when the velocity of the particle is 7 ms–¹.,t seconds after passing through O is given[br]by a = 4 - 2t.[/size][/size]

Solution[br][br](a) (i) Given acceleration function, [math]a=4-2t[/math][br]So, velocity function, [math]v=\int(4-2t)dt[/math][br][math]v=4t-t^2+c[/math][br]When [math]t=0[/math] and [math]v=4[/math],[br]So, [math]4=4(0)-0^2+c[/math][br] [math]c=4[/math][br]Thus, at time, [math]v=4t-t^2+4[/math] [br]When [math]t=7[/math], [math]v=4(7)-(7)^2+4[/math][br] [math]v=28-49+4[/math][br] [math]v=-17[/math][br][br]Hence, the instantaneous velocity of the particle when [math]t=7[/math] is [math]-17ms^{-1}[/math][br][br](ii) Maximum velocity,[math]ms^{-1}[/math][math](v)=0[/math][br]So, [math]4-2t=0[/math][br] [math]2t=4[/math][br] [math]t=2[/math] [br]Since [math]\frac{d^2v}{dt^2}=-2[/math] [math](<0)[/math][math]ms^{-1}[/math] , [math]v[/math] is maximum when [math]t=2[/math]. [br][br]Hence, maximum velocity of the particle [br] [math]=4(2)-(2)^2+4[/math] [br] [math]=8ms^{-1}[/math][br][br](b) When the instantaneous velocity of the particle is [math]7ms^{-1}[/math][br] [math]v=7[/math][br][math]4t-t^2+4=7[/math][br][math](t-1)(t-3)=0[/math] [math]t=1[/math] or [math]t=3[/math][br]Thus, the possible values of [math]t[/math]are 1 second and 3 seconds.