Six Questions To Test Kinematics.

Apply yourself

  1. Determine the absolute value of the velocity of a particle at time t=2s if the particle moves according to the law r=αt2i^+β(sinπt)j^\displaystyle r=\alpha { t }^{ 2 }\hat { i } +\beta (\sin { \pi t } )\hat { j } , with α=2m/s2\displaystyle \alpha =2{ m }/{ { s }^{ 2 } } and β=3m\displaystyle \beta =3m

  2. A particle moves according to the law r=αsin5ti^+β(cos25t)j^\displaystyle r=\alpha \sin { 5t } \hat { i } +\beta ({ cos }^{ 2 }5t)\hat { j } with α=2m\displaystyle \alpha =2m and β=3m\displaystyle \beta =3m.Find the velocity vector, acceleration vector, and trajectory of the particle's motion.

  3. The velocity of a particle varies according to the law v=α(2t3β)i^γ(sin2πt/3)j^\displaystyle v=\alpha (2{ t }^{ 3 }-\beta )\hat { i } -\gamma (\sin { 2\pi t/3 } )\hat { j } with α=1m/s4,β=1s3,γ=1m/s\displaystyle \alpha =1{ m }/{ { s }^{ 4 } },\quad \beta =1{ s }^{ 3 },\quad \gamma =1{ m }/{ s }. Find the law of motion if at the initial moment t=0 the particle was at origin.

  4. The acceleration of a particle varies according to the law a=αt2i^βj^\displaystyle a=\alpha { t }^{ 2 }\hat { i } -\beta \hat { j } with α=3m/s4\displaystyle \alpha =3{ m }/{ { s }^{ 4 } } and β=3m/s2\displaystyle \beta =3{ m }/{ { s }^{ 2 } }. Find the distance from the origin to the point where the particle will be at time t=1s if at t=0 the particle is at origin and has zero velocity.

  5. A train is moving rectilinearly with a velocity υ=180km/hr\displaystyle \upsilon =180{ km }/{ hr } . Suddenly an obstacle appears on the tracks and the brakes are applied.Now the velocity of the train changes according to ν=υαt2\displaystyle \nu =\upsilon -\alpha { t }^{ 2 } with α=1m/s2\displaystyle \alpha =1{ m }/{ { s }^{ 2 } } . What is the breaking distance of the train?How much time must pass from the moment the brakes are applied before the train stops?

  6. A rocket is launched from a land-based site vertically with an acceleration a=αt2\displaystyle a=\alpha { t }^{ 2 } , with α=1m/s4\displaystyle \alpha =1{ m }/{ { s }^{ 4 } } . At an altitude 100km above Earth's surface the rocket's boosters fail. How much time will pass (from the moment the boosters failed) before the rocket crashes? Air drag is ignored. The initial velocity is 0.

Don't forget to explain yourself in comments.

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Note by Soumo Mukherjee
6 years, 9 months ago

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Q1. v=drdt=2αti^+βπcos(πt)j^=8i^+3πj^\vec{v}=\frac{d\vec{r}}{dt}=2\alpha t \hat{i}+\beta\pi cos(\pi t) \hat{j}=8 \hat{i}+3\pi \hat{j}

v=12.3622\Rightarrow |\vec{v}|=\boxed{12.3622}

Q2. v=drdt=5αcos(5t)i^5βsin(10t)j^\vec{v}=\frac{d\vec{r}}{dt}=\boxed{5\alpha cos (5t)\hat{i}-5\beta sin(10t) \hat{j}}

a=dvdt=25αsin(5t)i^50cos(10t)j^\vec{a}=\frac{d\vec{v}}{dt}=\boxed{-25\alpha sin (5t) \hat{i} -50 cos (10t) \hat{j}}

x=αsin(5t)x2α2=sin2(5t);yβ=cos2(5t)x=\alpha sin (5t)\Rightarrow \frac{x^{2}}{\alpha^{2}}=sin^{2}(5t); \frac{y}{\beta}=cos^{2}(5t)

So x2α2+yβ=1\boxed{\frac{x^{2}}{\alpha^{2}}+\frac{y}{\beta}=1} (A Parabola)

Q6. a=αt2a=\alpha t^{2}

0vdv=0tαt2dt\displaystyle\int^v_0 dv=\displaystyle\int^t_0 \alpha t^{2} dt v=αt33\Rightarrow v=\alpha\frac{t^{3}}{3}

0xdx=0tαt33dt\displaystyle\int^x_0 dx=\displaystyle\int^t_0 \alpha \frac{t^{3}}{3} dt x=αt412\Rightarrow x=\alpha\frac{t^{4}}{12}

So time when rocket will be at a height of 100 km is (12×105)14 (12×10^{5})^{\frac{1}{4}}=33.0975 seconds

So velocity at this time will be 12085.5016 m/s

Assuming constant gravity and using 2nd2^{nd} equation of motion, one can find the time! I wont be doing it now as Latex is necessary evil (atleast after typing so much!)

@Math Philic you typoed in Q5! I think velocity of train must change in such clean weather (and not rain)

Pranjal Jain - 6 years, 9 months ago

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Oh!train and rain! thanx for pointing it out.

Q2.:The trajectory you found has a standard name.Don't you think mentioning it would be better.

Q6: I suggest you to go through it once more.

All the questions are such that to provoke analysis of solutions after solving the problems.Yes, you need to dig deeper even after shooting your enemy dead.So the problems are really solved when you have explicit reasons to believe the whole solution is correct.Are they really dead?Well, there are cases of Zombies now a days.Go through it again.

Nice work @Pranjal Jain .Try more questions and be the first to complete all of them.Such questions come in competitive exams.So you will benefit from them.

Soumo Mukherjee - 6 years, 9 months ago

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good questions

Ashish Sharma - 5 years, 10 months ago

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