Question

please solve question no. 1

Answer 1

ANSWER:-
Notice that we have a right triangle with side lengths x(t)−vΔtx(t)−vΔt, vΔtvΔt, and x(t+Δt)x(t+Δt). Using the pythagorean theorem, we obtain the equation:
x(t+Δt)2=(x(t)−vΔt)2+(vΔt)2x(t+Δt)2=(x(t)−vΔt)2+(vΔt)2
Expanding the squared equations we get:
x(t+Δt)2=x(t)2−2x(t)vΔt+2(vΔt)2x(t+Δt)2=x(t)2−2x(t)vΔt+2(vΔt)2
Now, let α(t)=x(t)2α(t)=x(t)2:
α(t+Δt)=α(t)−2α(t)−−−−√vΔt+2(vΔt)2α(t+Δt)=α(t)−2α(t)vΔt+2(vΔt)2
More algebra, move α(t)α(t) over, divide by ΔtΔt
α(t+Δt)−α(t)Δt=−2α(t)−−−−√v+2v2Δtα(t+Δt)−α(t)Δt=−2α(t)v+2v2Δt
Now notice if we let ΔtΔt go to 0 (which represents continuous motion), the left side is just a derivative:
dα(t)dt=−2α(t)−−−−√vdα(t)dt=−2α(t)v
Now this is just a differential equation that is easy to solve with separation of variables:
dαα(t)√=−2vdtdαα(t)=−2vdt
Note, our integral over α(t)α(t) will go from d2d2 to x(t)2x(t)2.
∫x(t)2d2dα(t)α(t)√=∫t0−2vdt∫d2x(t)2dα(t)α(t)=∫0t−2vdt
(2α(t)−−−−√)x(t)2d2=−2vt(2α(t))d2x(t)2=−2vt
|x|−|d|=−vt|x|−|d|=−vt
Now we can solve for t when x=0x=0 (d is assumed to be positive, so we drop the absolute value):
t=dvt=dv.

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