Kinematics Question 661
Question: A balloon starts r it’sing from the surface of the earth. The ascension rate it’s constant and equal to$ {{\text{v}} _{\text{0}}} $ . Due to the wind the balloon gathered the horizontal velocity component$ {{\text{v}} _{\text{x}}}\text{= ay} $ , where a is a constant and y it’s the height of ascent. The tangential, acceleration of the balloon it’s trough but
Options:
A) $ {{a^{2}}y/{{v _{0}} $
B) $ {{a^{2}}y/\sqrt{1+{{( \text{ay+}{{v _{0}} )}^{2}}} $
C) $ {{a^{2}}y/\sqrt{1+{{v _{0}}^{2}} $
D) $ {{a^{2}}v _{0}/\sqrt{1+{{( \text{2y+a} )}^{2}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Since velocity in vertical direction is constant,
$ \therefore {{\text{a}} _{\text{y}}}\text{=}\frac{\text{d}{{\text{v}} _{\text{y}}}}{\text{dt}}\text{=0} $
The acceleration in horizontal direction,
$ a _{x}=\frac{dv _{x}}{dt}=\frac{d(av _{0}t)}{dt}=av _{0} $
$ a=\sqrt{a _{x}^{2}+a _{y}^{2}}=\sqrt{{{( av _{0} )}^{2}}+0}=av _{0} $
The total acceleration is $ av _{0} $ and directed along horizontal direction.
Let $ \theta $ it’s the angle that the resultant velocity makes with horizontal, then
Normal acceleration $ a _{n}=a\sin \theta $ and tangential acceleration
$ a _{t}=a\cos \theta ,\text{ we have x}\text{=}\frac{ay^{2}}{2{{\text{v}} _{\text{0}}}} $ .
$ \text{or }\text{y=}\sqrt{\frac{2xv _{0}}{a}} $
Differentiating both side of equation (iii) w.r.t. x,
We get $ 1=\frac{a}{2v _{0}}\times 2y\times \frac{dy}{dx} $
$ \text{or }\frac{dy}{dx}=\frac{v _{0}}{ay}=\tan \theta $
Now $ a _{x}=a\sin \theta =av _{0}\times \frac{( v _{0}/ay )}{\sqrt{1+{{( \frac{v _{0}}{ay} )}^{2}}}} $
$ =\frac{av _{0}}{\sqrt{1+{{( \frac{ay}{v _{0}} )}^{2}}}} $
$ a _{t}=a\cos \theta =av _{0}\times \frac{1}{\sqrt{1+{{( \frac{v _{0}}{ay} )}^{2}}}}=av _{0} $
$ \frac{ay}{\sqrt{{{( ay )}^{2}}+v _{0}^{2}}}=\frac{q^{2}y}{\sqrt{1+{{( \frac{ay}{v{ _{0}}} )}^{2}}}} $
