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Classical Mechanics Level 4

Given, the position vector of a particle as

r ⃗ = ( 3 sin ⁡ t + 25 ) i ⃗ − ( 3 cos ⁡ t − 45 ) j ⃗ + 24 k ⃗ \vec r = (3 \sin t + 25)\vec i - (3 \cos t -45) \vec j + 24\vec k metres.

Where t t is time in seconds..

Find the distance travelled by the particle in time from t = 1999 sec t=1999\text{ sec} to t = 2016 sec t=2016 \text{ sec} ... If it can be represented as a a , then find a 2 a^2 .

A really cool problem...tests you knowledge....of B a s i c s Basics

This is a part of my set Aniket's Mechanics Challenges


The answer is 2601.0.

Aniket Sanghi by Aniket Sanghi , 21, India

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2 solutions

Otto Bretscher
Mar 6, 2016

No calculus required: Since we are going round a circle of radius 3, with a constant speed of 3, we are traveling a = 3 ∗ ( 2016 − 1999 ) a=3*(2016-1999) units, and the answer is a 2 = 5 1 2 = 2601 a^2=51^2=\boxed{2601}

5 Helpful 0 Interesting 2 Brilliant 0 Confused
Aniket Sanghi
Mar 5, 2016

Start by differentiating and finding velocity vector which comes out to be v = ( 3 c o s t ) i + ( 3 s i n t ) j v=(3 cost)i + (3 sint)j

.now we will find the speed of particle |v| which comes out to be constant and is equal to |v|=3m/sec

now time change is 17 seconds so distance travelled is 17× 3=51 (a) .... 5 1 2 51^2 = 2601.0 \boxed {=2601.0 }

4 Helpful 0 Interesting 0 Brilliant 0 Confused

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