Answer:
The particle changes its direction 2 times within the time -3<t<3
Step-by-step explanation:
The particle is moving only in a single dimension (x-axis), and whenever the particle changes its direction it also means that it's velocity while changing the direction will be zero.
Hence,
v(t) = 0
but since we're not concerned with the actual values of t when v(t)=0, we'll only consider how many times does the particle changes its direction.
for that we'll simply plot the curve using half-steps from -3 to 3.
t, v(t)
-3, 115
-2.5, 23.9375
-2, -16
-1.5, -25.0625
-1, -19
-0.5, -9.0625
0, -2
0.5, -0.0625
1, -1
1.5, 1.9375
2, 20
2.5, 68.9375
3, 169
What we need to check is at what points does the sign of v(t) values change (because only between those points will v(t) cross the x-axis, hence it's value would've crossed 0)
so there are two points!
between the intervals t = [-2.5,2] and [1,1.5]
so there are two points where the particle changes its directions and those points lie somewhere between these two aforementioned intervals.
Yes and no it depends on what the answer is
Step-by-step explanation:
Solve for a
k + a = v + w
a = v + w - k
Solution → { a = v + w - k }
(5+1)2 - (12+32) + 3
6 (2) - 44 + 3
12 - 44 + 3 = -29
Wow that is hardddddddddddddddddddddddddddddddddddddddd
