Determine the absolute value of the velocity of a particle at time t=2s if the particle moves according to the law , with and
A particle moves according to the law with and .Find the velocity vector, acceleration vector, and trajectory of the particle's motion.
The velocity of a particle varies according to the law with . Find the law of motion if at the initial moment t=0 the particle was at origin.
The acceleration of a particle varies according to the law with and . 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.
A train is moving rectilinearly with a velocity . Suddenly an obstacle appears on the tracks and the brakes are applied.Now the velocity of the train changes according to with . What is the breaking distance of the train?How much time must pass from the moment the brakes are applied before the train stops?
A rocket is launched from a land-based site vertically with an acceleration , with . 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.
Have a nice day.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Q1. v=dtdr=2αti^+βπcos(πt)j^=8i^+3πj^
⇒∣v∣=12.3622
Q2. v=dtdr=5αcos(5t)i^−5βsin(10t)j^
a=dtdv=−25αsin(5t)i^−50cos(10t)j^
x=αsin(5t)⇒α2x2=sin2(5t);βy=cos2(5t)
So α2x2+βy=1 (A Parabola)
Q6. a=αt2
∫0vdv=∫0tαt2dt ⇒v=α3t3
∫0xdx=∫0tα3t3dt ⇒x=α12t4
So time when rocket will be at a height of 100 km is (12×105)41=33.0975 seconds
So velocity at this time will be 12085.5016 m/s
Assuming constant gravity and using 2nd 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)
Log in to reply
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.
good questions