At time t=0, the velocity of a particle moving along a straight line is v(t)=t^2-4t+3.

a) Find the particle's acceleration each time the velocity is zero.
b) When is the particle moving forward?
c) When is the particle's velocity decreasing

Question

At time t>=0, the velocity of a particle moving along a straight line is v(t)=t^2-4t+3.

a) Find the particle's acceleration each time the velocity is zero.
b) When is the particle moving forward?
c) When is the particle's velocity decreasing

anonymous · Answer

wish i could read, that is the velocity.
ok interpret the first question as set 
$$t^2-4t+3=0$$ solve for $t$

anonymous · Answer

easy for this one since it factors as 
$$(t-1)(t-3)=0$$

anonymous · Answer

2) moving forward means $v(t)>0$ and since this is a parabola that opens up, it is negative between the roots and positive outside them

anonymous · Answer

3) take the derivative, get a line. solve for where the line is negative, aka below the $x$ axis

anonymous · Answer

So at t=1, t=3 is the acceleration?

anonymous · Answer

no, that is where the velocity is zero
that is the answer to question 1

anonymous · Answer

oh man i really can't read jesus, ignore everything i wrote 
it is asking about ACCELERATION !  sorry, let me delete everything, put on a dunce cap and start again

anonymous · Answer

haha its okay

anonymous · Answer

velocity is $v(t)=t^2-4t+3$ and it is zero at $t=1,t=3$ at least we got that right

anonymous · Answer

would I just plug in V(1) and v(3)?

anonymous · Answer

acceleration  is $a(t)=2t-4$ and the first question asks what is acceleration when the velocity is $0$ to take $a(1)$ and $a(3)$

anonymous · Answer

plug in to the derivative for acceleration
if you plug back in to $v$ you get zero, because that is how you solved for $1$ and $3$ to begin with

anonymous · Answer

oh okay

anonymous · Answer

now for question two my answer was actually correct, because it asks when it is moving forward and  backward.  
$v(t)=t^2-4t+3$ is a parabola facing up so it is negative between $1$ and $3$ and positive outside , i.e. when $t<1$ and when $t>3$

anonymous · Answer

for question 3, velocity is decreasing when acceleration is negative, so solve 
$$2t-4<0$$ which should be easy enough
sorry i messed up so bad, my reading comprehension is not what it should be

anonymous · Answer

its okay. Thanks a lot! Im going to solve it out and let you know what I get