Question

Linear motion with variable forces

Answer 1

A particle is at the origin at time t=0, and its velocity is given by the following equations. Find equations containing v with x, and find expressions for the acceleration.
(i) in terms of x, (ii) in terms of v.

\[\text{For} ~~~v=\sec^2(t) ~~~\text{at} ~~~0 \le t \le \frac{\pi}{2}\]

Answer 2

what I did is
\[\frac{dv}{dt}=\sec^2(t)\tan(t) ~~~~~||~~~~~ \frac{dx}{dt}=\sec^2(t)\]
\[\int\limits 1 dx =\int\limits \sec^2(t)dt \\ \\ x=\tan(t)\]
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\[\frac{dv}{dt}=\frac{dv}{dx} \times \frac{dx}{dt} \\ \\ \frac{dv}{dx}=\frac{\sec^2(t)\tan(t)}{\tan(t)} \\ \\ \frac{dv}{dx}=\sec^2(t) \\ \\ \int\limits 1dv = \int\limits \sec^2(t) dx \\ \\ v=xsec^2(t)\]
can't solve it
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\[Ans:~~~ v=1+x^2 ~~~~~||~~~~~2x(1+x^2)~~~~~||~~~~~2v \sqrt{v-1}\]

Answer 3

@Jemurray3

Answer 4

I studied physics in college algebra based, I don't know how to apply calculus to physics

Answer 5

Just take the antiderivative of sec^2(t), which is tan(t).

Answer 6

you were saying
\[v=\sec^2t \\ \\ \frac{dx}{dt}=\sec^2t \\ \\ \int\limits 1dx = \int\limits \sec^2tdt \\ \\ x=\tan(t)+k\]
is it?

Answer 7

@Peter14

Answer 8

what?

Answer 9

do u know how to solve this?

Answer 10

maybe, it's been awhile since I took calculus.

Answer 11

I studied algebra based physics in college, I don't know how to apply calculus to physics

Answer 12

I don't quite understand what it means "in terms of x"

Answer 13

probably it says find accleration in terms of x, like a=x^2

Answer 14

oh. for in terms of x you are supposed to relate the acceleration to the position.

Answer 15

and for in terms of v you are supposed to relate the acceleration to the velocity

Answer 16

so take the integral of sec^2 t for the POSITION (I got this completely wrong earlier, I took it the wrong direction) and the derivative of sec^2 t for the ACCELERATION

Answer 17

the integral is simple, you get tan t + C and they give you C=0 at the beginning of the problem.
so the position (x) of the particle is given by x=tan t
The derivative is more complicated, you use the product rule of differentiation to get
\[\frac{ d }{ dt }\sec^2t=\frac{ d }{ dt }(\sec t)(\sec t)=(\sec t) (\sec t \tan t)+ (\sec t)(\sec t \tan t)=2\sec^2 t \tan t\]
so the acceleration of the particle is given by\[\alpha=2\sec^2 t \tan t\]

Answer 18

to find the acceleration relative to the position and the velocity you have to use trig inverses to find how t varies with x and v.

Answer 19

so solve the equations for velocity and position for t, then plug the ts you get into the equation for acceleration.

Answer 20

so you get

x = tan (t) + k
and since particle was at rest at t = 0 .. you get k = 0
hence
x= tan(t)
and v = sec^2(t)..
v= tan^2(t) + 1
so v= x^2 +1 .. tada :D

Answer 21

where did u get v= tan^2(t) + 1

Answer 22

OH I SEE

Answer 23

from trig identities

Answer 24

yup.. good old trig