Mathematics, 21.12.2020 17:30, memos5502

Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units of m/s. Complete parts a through c. a. Determine when the motion is in the positive direction and when it is in the negative direction.
b. Find the displacement over the given interval.
c. Find the distance traveled over the given interval.
v(t) = 3t^2 - 24t + 36; [0.7]

Answers: 3

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answered: shortiibrownskiin

a. the motion is positive in the time intervals: $[0,2)U(6,\infty)$

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:

$3t^{2}-24t+36=0$

so let's solve this for t:

$3(t^{2}-8t+12)=0$

$t^{2}+8t+12=0$

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:

$[0,2) (2,6) (6,\infty)$

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:

$v(1)=3(1)^{2}-24(1)+36$

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:

$v(1)=3(3)^{2}-24(3)+36$

v(3)=-9 m/s

we got a negative value so the object moves in the negative direction.

$(6,\infty)$ test value t=7

so:

$v(1)=3(7)^{2}-24(7)+36$

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: $[0,2)U(6,\infty)$

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:

$s(t)=\int\limits^7_0 {(3t^{2}-24t+36)} \, dt$

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{0}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{7}_{0}$

so now we evaluate the integral:

$s=7^{3}-12(7)^{2}+36(7)-(0^{3}-12(0)^{2}+36(0))$

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:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{2}_{0}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{2}_{0}$

so now we evaluate the integral:

$s=2^{3}-12(2)^{2}+36(2)-(0^{3}-12(0)^{2}+36(0))$

s=32 m

(2,6)

s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{6}_{2}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{6}_{2}$

so now we evaluate the integral:

$s=6^{3}-12(6)^{2}+36(6)-(2^{3}-12(2)^{2}+36(2))$

s=-32 m

(6,7)

s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{6}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{7}_{6}$

so now we evaluate the integral:

$s=7^{3}-12(7)^{2}+36(7)-(6^{3}-12(6)^{2}+36(6))$

s=7 m

and we now add all the magnitudes:

Distance=32+32+7=71m

answered: Guest

y = 0.77(x)

step-by-step explanation:

the jacket retains this value after 23% cut or retains 77% of original value. plug original 87 jacket price for x and the remaining value should be y.

answered: Guest

Look, when you want to do something like ${x}^{2}$ put it like x^2. x2 could mean x² or x₂. so to clear confusion, know the notation. to find the product multiply the x's and you get -8x and the y's and you get 6y. since it's all being multiplied it is just -8x6y, or answer a. although the better answer is -48xy(combining -8 and 6) because you should multiply constants whenever you can.

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Consider an object moving along a line with the given velocity v. Assume t is time measured in secon...