Third-degree, with zeros of -3,-2, and 1, and passes through the point (3,9)
1 answer:
You might be interested in
Answer:
a. the motion is positive in the time intervals:
The motion is negative in the time interval: (2,6)
b. S=7 m
c. distance=71m
Step-by-step explanation:
a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:
so let's solve this for t:
and now we factor it again:
(t-6)(t-2)=0
so we get the following answers:
t=6 and t=2
so now we can build our possible intervals:
and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:
[0,2) test value t=1
so:
v(1)=15 m/s
we got a positive value so the object moves in the positive direction.
(2,6) test value t=3
so:
v(3)=-9 m/s
we got a negative value so the object moves in the negative direction.
test value t=7
so:
v(1)=15 m/s
we got a positive value so the object moves in the positive direction.
the motion is positive in the time intervals:
The motion is negative in the time interval: (2,6)
b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:
so we get:
which simplifies to:
so now we evaluate the integral:
s=7 m
for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:
[0,2)
s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt
so we get:
which simplifies to:
so now we evaluate the integral:
s=32 m
(2,6)
s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt
so we get:
which simplifies to:
so now we evaluate the integral:
s=-32 m
(6,7)
s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt
so we get:
which simplifies to:
so now we evaluate the integral:
s=7 m
and we now add all the magnitudes:
Distance=32+32+7=71m