# Word problem

The function s(t)=t2+tlnt gives the position from time [1,e] find:

a. the velocity at time t
b. the acceleration at time t
c. describe the motion of the object: when is it moving forward or moving backward. also compute the total distance traveled during the interval
d. when is the object accelerating and when is it decellerating
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my work
s(t)=t2+tlnt
v(t)=2t+1+lnt
a(t)=2+(1/t)

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Are my answers for a and b correct? And how would I find parts c and d.

Thank you!

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Mark44
Mentor
The function s(t)=t2+tlnt gives the position from time [1,e] find:

a. the velocity at time t
b. the acceleration at time t
c. describe the motion of the object: when is it moving forward or moving backward. also compute the total distance traveled during the interval
d. when is the object accelerating and when is it decellerating
------------------------------------------------------------

my work
s(t)=t2+tlnt
v(t)=2t+1+lnt
a(t)=2+(1/t)

-------------------
Are my answers for a and b correct? And how would I find parts c and d.

Thank you!
a and b are fine.
For c, the object is moving forward when v(t) > 0, and backward when v(t) < 0.
For d, the object is accelerating when its acceleration is positive; i.e., when a(t) > 0. The object is decelerating when a(t) < 0. Keep the function's domain in mind when you answer these parts.

a and b are fine.
For c, the object is moving forward when v(t) > 0, and backward when v(t) < 0.
For d, the object is accelerating when its acceleration is positive; i.e., when a(t) > 0. The object is decelerating when a(t) < 0. Keep the function's domain in mind when you answer these parts.
2t+1+lnt>0 and 2t+1+lnt<0

Howw would you even solve that inequality?

2+(1/t)>0
1/t>-2
-1/2<t
So its accelerating at t>-1/2 and declerating at t<-1/2?
Would I even write when that since those numbers arent in the domain [1,e]?

Ray Vickson
Homework Helper
Dearly Missed
2t+1+lnt>0 and 2t+1+lnt<0

Howw would you even solve that inequality?

2+(1/t)>0
1/t>-2
-1/2<t
So its accelerating at t>-1/2 and declerating at t<-1/2?
Would I even write when that since those numbers arent in the domain [1,e]?
Are numbers in [1,e] greater than -1/2 or less than -1/2?

RGV

So the answer for part d would be that the object accelerates between [1,e] but does not decelerate?

can you show me how to do part c, I have no idea how to do that part

Mark44
Mentor
So the answer for part d would be that the object accelerates between [1,e] but does not decelerate?
Yes. There is no deceleration because acceleration >= on the interval [1, e].
can you show me how to do part c, I have no idea how to do that part
For c, you have v(t) = 2t + 1 + ln(t). You will not be able to solve the inequality 2t + 1 + ln(t) > 0, but since t ε [1, e], you should be able to determine the sign of the individual terms to say something about the sign of the sum of these terms.

For the other part of c, integrate the velocity over the interval [1, e] to get the total displacement.

Yes. There is no deceleration because acceleration >= on the interval [1, e].
For c, you have v(t) = 2t + 1 + ln(t). You will not be able to solve the inequality 2t + 1 + ln(t) > 0, but since t ε [1, e], you should be able to determine the sign of the individual terms to say something about the sign of the sum of these terms.

For the other part of c, integrate the velocity over the interval [1, e] to get the total displacement.
So the object is moving forward

s(1)=t2+tlnt
s(1)=1

s(e)=e2+elne
s(e)=10.1017

10.107-1=9.107

it goes 9.107 units foward??

Mark44
Mentor
So the object is moving forward

s(1)=t2+tlnt
s(1)=1

s(e)=e2+elne
s(e)=10.1017

10.107-1=9.107

it goes 9.107 units foward??
That's different from what I said, but both techniques produce the same result. The exact distance travelled is e2 + e - 1, which is approximately 10.107 units.

That's different from what I said, but both techniques produce the same result. The exact distance travelled is e2 + e - 1, which is approximately 10.107 units.
Alright thanks. WE haven't learned integration yet do I couldn't do it your way