Question

Find the value of t when the particle is instantaneously at rest. What does this question mean? (image attached)

Answer 1

the particle is instantaneously at rest when its velocity is zero

Answer 2

so solve for \(v(t)=0\); focus on each piece separately: $$5-(t-2)^2=0\\(t-2)^2=5\\t-2=\pm\sqrt5\\t=2\pm\sqrt5$$now, \(\sqrt5>\sqrt4=2\) so \(2-\sqrt5<0\) and \(2+\sqrt5>4\), but this piece of the velocity function is only for \(0\le t\le4\), so neither of these solutions are valid

Answer 3

now, what for \(t>4\)? we solve:$$3-\frac{t}2=0\\\frac{t}2=3\\t=6$$and since \(6>4\) this solution is actually valid and \(t=6\) is where our particle is instantaneously at rest

Answer 4

@oldrin.bataku that's what I don't understand: why is the solution not valid?

Answer 5

@freshfaced \(v(t)=5-(t-2)^2\) only for \(0\le t\le4\). when we solved \(5-(t-2)^2=0\) we found solutions \(t\) that were not in this interval and thus are not solutions to \(v(t)=0\)

Answer 6

ohh ok. I see, so only the second equation is valid?

Answer 7

how would you do part 2 then? (attached)

Answer 8

for one, \(v(t)\) is undefined for \(t<0\) so the first solution is garbage; for the second solution \(t=2+\sqrt5\), notice \(2+\sqrt5>4\) so when evaluating \(v(2+\sqrt5)\) we'd actually be using the *second* branch (\(3-t/2\)) rather than the first (\(5-(t-2)^2\)). this means \(v(2+\sqrt5)=3-(2+\sqrt5)/2=1-\sqrt5/2\ne0\)

Answer 9

as well, is there significance in the small 't' and the big 'T'?

Answer 10

@oldrin.bataku for part 2

Answer 11

well, presumably the initial position is at \(t=0\). they're saying at some later time \(t=T\) the particle will return there (\(T\) is a constant we're solving for that denotes the specific time at which the particle returns to its initial position while \(t\) is a variable denoting time)

Answer 12

now, recall that displacement is the integral of velocity: \(\Delta x=\int_{t_0}^{t_1} v(t)\ dt\) so for the particle to return to the origin we expect \(\Delta x=0\) and the integral to be over the interval \((0,T)\): $$\int_0^T v(t)\ dt=0$$

Answer 13

now, we have two possibilities: \(0\le T\le 4\), in which case our entire integral is in the first branch; or \(T>4\), in which we'll have an integral that spans over both branches which we'll have to break into two integrals

Answer 14

ok

Answer 15

@oldrin.bataku I found the d=47/3, but that is still not the answer?

Answer 16

in the first case, \(0\le T\le 4\), we can substitute \(v(t)=5-(t-2)^2\) so:$$\int_0^T (5-(t-2)^2)\ dt=\int_0^T (5-(t^2-4t+4))\ dt=\int_0^T (1+4t-t^2)\ dt$$integrating term by term, we have: $$\int_0^T(1+4t-t^2)\ dt=\left[t+2t^2-\frac13t^3\right]_0^T=T+2T^2-\frac13T^3$$hence we end up solving: \(T+2T^2-\frac13T^3=0\\T(1+2T-\frac13T^2)=0\\T(T^2-6T-3)=0\)
... which yields *is* the initial point of the motion rather than a return to the initial position. with \(3+2\sqrt3>4\) and \(3-2\sqrt3<0\) they are invalid solutions on the interval, so now we focus our attention on \(T>4\)

Answer 17

ok so solving the one with T>4 and equating it to d right?

Answer 18

$$\int_0^T v(t)\ dt=\int_0^4(5-(t-2)^2)\ dt+\int_4^T (3-\frac{t}2)\ dt$$in the first integral, we already evaluated that integral for upper bound \(T\) to be \(T+2T^2-\frac13T^3\) above so evaluating with \(T=4\) we get \(4+2\cdot16-\frac13\cdot64=\frac{44}3\)

now, we'll evaluate hte second integral: $$\int_4^T\left(3-\frac{t}2\right)\ dt=\left[3t-\frac14t^2\right]_4^T=3T-\frac14T^2-12+\frac14\cdot4^2=3T-\frac14T^2-8$$

Answer 19

so now we have:$$\int_0^T v(t)\ dt=0\\\frac{44}3+3T-\frac14T^2-8=0\\\frac{20}3+3T-\frac14T^2=0\\T^2-12T-\frac{80}3=0$$

Answer 20

ok

Answer 21

this gives $$T=6\pm2\sqrt{\frac{47}3}$$now we notice that since we're interested in \(T>4\), only the second solution \(T=6+2\sqrt{\frac{47}3}\) is valid and this is the time at which our particle returns to its original position

Answer 22

oh wow this is really confusing. thank you!