A particle moves in a straight line such that its velocity (in m/s) is given by v=4t^2-2t+3 where t is time in seconds. Fin
a) The velocity when t =2,
V= 11m/s
This first part is just arithmetic: 4(2^2)- 2(2)+ 3= 16- 4+ 3= 12+ 3= 15 m/s, not 11 m/s.
b) the acceleration function,
is it? a= 8t-2?
I am puzzled as to why this is posted under "Beginning Algebra". You
cannot do part (b) without Calculus. If the velocity at any time, t, is given by 4t^2- 2t+ 3, then the acceleration is given by dv/dt= 8t- 2.
c) the value of t for which the particle has maximum velocity
I dont know how to start to make this question and I would like to know if part a and b are right.
Are you sure you have copied the problem correctly? The graph of this function is a parabola that opens
upward. It does NOT have a "maximum". I would suggest that either you were supposed to have y= -4t^2- 2t+ 3 or you are to find the
minimum velocity.
If you are to find the minimum velocity, there are two ways to do . As long as the acceleration is positive, the velocity is already increasing and cannot be minimum. If the acceleration is negative, the velocity is still decreasing and so is not the minimum. Set the acceleration function equal to 0 and solve for t.
Or complete the square: v= 4t^2- 2t+ 3= 4(t^2- (1/2)t+ 1/16- 1/16)= 4(t^2- (1/2)+ 1/16)- 1/4= 4(t- 1/4)^2- 1/4. If t= 1/4, t- 1/4= 0 so this is just -1/4. If t is any other number, (t- 1/4)^2 is positive and we have v greater than -1/4. We have minimum v, -1/4, for t= 1/4.
