1. ## Displacement and Distance

A particle moves with acceleration a(t)m/sec^2 along the s-axis and has velocity Vm/sec at time t=0. Find the displacement and distance traveled by the particle during the given time interval.

a(t)=-2;V=3;[1,4]

I got the answer 9m for both displacement and distance...is this correct?

2. Hello, holly123!

A particle moves with acceleration: $a(t) \,=\,2$ along the s-axis
. . and has velocity $v = 3$ when $t=0.$

Find the displacement and distance traveled during the interval $[-4,1]$

We have: . $a(t) \:=\:2$

Integrate: . $v(t) \:=\:2t + C_1$

When $t=0,\:v = 3 \;\hdots \text{so we have: }\:3 \:=\:2(0) + C_1 \quad\Rightarrow\quad C_1 = 3$

. . Hence: . $v(t) \:=\:2t+3$

Integrate: . $(t) \:=\:t^2 + 3t + C$

When $t = \text{-}4\!:\;\;s(\text{-}4) \:=\:(\text{-}4)^2 + 3(\text{-}4) + C \:=\: C + 4$
When $t=1\!:\;\;s(1) \:=\:1^2 + 3(1) + C \:=\:C+4$

At the beginning of the interval, the particle is at $C + 4.$
At the end of the interval, the particle is still at $C + 4.$

. . $\boxed{\text{The displacement is 0 (zero).}}$

Set the velocity equal to zero: . $2t + 3 \:=\:0 \quad\Rightarrow\quad t \:=\:-\tfrac{3}{2}$
This is when the particle stops and changes direction.

When $t = \text{-}4\!:\;\;s(\text{-}4) \:=\:C+4$
When $t = \text{-}\tfrac{3}{2}\!:\;\;s\left(\text{-}\tfrac{3}{2}\right) \:=\:C - \tfrac{9}{4}$
On the interval $\left[\text{-}4,\,\text{-}\tfrac{3}{2}\right]$, it moved: . $\left(C - \tfrac{9}{4}\right) - \left(C + 4\right) \:=\:-\tfrac{25}{4}$
. . It moved $\tfrac{25}{4}$ units to the left.

When $t = \text{-}\tfrac{3}{2}\!:\;\;s\left(\text{-}\tfrac{3}{2}\right) \:=\:C - \tfrac{9}{4}$
When $t = 1\!:\;\;s(1) \:=\:C + 4$
On the interval $\left[\text{-}\tfrac{3}{2},\:1\right]$, it moved: . $(C+4) - \left(C - \tfrac{98}{4}\right) \:=\:\tfrac{25}{4}$
. . It moved $\tfrac{25}{4}$ units to the right.

Its total distance traveled is: . $\frac{25}{4} + \frac{25}{4} \:=\:\boxed{\frac{25}{2}}$ units.

3. a(t)= -2 though....do negatives not matter for acceleration?