A classical mechanics problem by Rishabh Tiwari

A particle is moving along x x -axis with a constant acceleration of 2 ms 2 -2 \text{ ms}^{-2} . The particle leaves the origin with a velocity of 5 ms 1 5 \text{ ms}^{-1} . Determine the distance (in m \text{m} ) travelled by the particle during the third second of its motion.


The answer is 0.5.

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4 solutions

Sabhrant Sachan
Jun 12, 2016

Notice that the particle have initial velocity towards + x + x axis and acceleration towards x - x axis

If we apply the direct formula to calculate the Displacement in n t h n^{th} second , we get

S n = u + a 2 ( 2 n 1 ) S 3 = 5 2 2 ( 5 ) = 0 S_{n}=u+\dfrac{a}{2}(2n-1) \implies S_{3}=5-\dfrac{2}{2}\cdot(5) = 0

Now to calculate the distance , we have to use laws of motion

First Let's calculate the time when velocity of the particle is 0 0

v = u + a t 0 = 5 2 t t = 2.5 Seconds v=u+at \implies 0=5-2\cdot t \\ t = 2.5 \text{ Seconds }

Total Distance traveled by the particle ( S ) (S) will be

S = 2 ( S 2.5 S 2 ) S = 2 ( 5 2.5 1 2 2 6.25 5 2 + 1 2 2 4 ) S = 2 ( 12.5 6.25 10 + 4 ) S = 0.5 m S = 2(S_{2.5}-S_{2}) \\ S = 2\left(5\cdot 2.5-\dfrac{1}{\cancel{2}}\cdot\cancel{2}\cdot6.25-5\cdot 2+\dfrac{1}{\cancel{2}}\cdot\cancel{2}\cdot4 \right) \\ S= 2 (12.5-6.25-10+4) \\ \boxed{S=0.5} \text{ m }


Nice problem , i loved it :)

Thank you :-) Nice solution (+1)! You loved it but you didn't liked it :(

Rishabh Tiwari - 5 years ago
Raj Rajput
Jun 12, 2016

Nice approach! (+1)!

Rishabh Tiwari - 5 years ago

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Thanks :):)

RAJ RAJPUT - 5 years ago

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Did you appeared for jee this year or you are in 11th ?

Rishabh Tiwari - 5 years ago
Rishabh Tiwari
Jun 12, 2016

Let's see when the particle comes to rest ,

From v = u + at \text{v = u + at} ,

0 0 = = 5 5 - 2 t 2t

\Rightarrow t t = = 2.5 s 2.5 s ,

Hence, the particle changes its direction of motion at t t = = 2.5 s 2.5s , so we cannot directly use the formula of distance travelled in n t h n^{th} sec. Therefore we will calculate distance travelled from t = 2 t=2 to t = 2.5 t=2.5 & from t = 2.5 t=2.5 to t = 3 t=3 separately !

Now , we use x x = = u t ut + + 1 2 \dfrac{1}{2} a a t 2 t^{2}

For displacement in t = 2 t=2 to t = 2.5 t=2.5 , we subtract the displacement at t = 2 t=2 from displacement at t = 2.5 t=2.5 :-

x 1 x_1 = = 5 ( 2 ) 5\cdot(2) - 4 4 \Rightarrow x 1 x_1 = = 6 m 6m ....(disp. in t = 2s)

x 2 x_2 = = 5 ( 2.5 ) 5\cdot(2.5) - 6.25 6.25 \Rightarrow x 2 x_2 = = 6.25 m 6.25m ....(disp. in t = 2.5s)

Required displacement = = x 2 x 1 x_2 - x_1 = = 0.25 m 0.25m .......................... ( 1 ) (1)

Now our half work is done, For t = 2.5 t=2.5 to t = 3 t=3 , we have :

u u = = 0 0

a a = = + 2 +2

t t = = 0.5 0.5

Putting these values in

x x = = u t ut + + 1 2 \dfrac{1}{2} a a t 2 t^{2}

We get x x = = 0.25 m 0.25m ...................... ( 2 ) (2)

Adding ( 1 ) (1) & ( 2 ) (2) we get net displacement in 3rd second as x = 0.5 m \boxed{x=0.5m} .

Please feel free to comment & correct me if I am wrong & would really appreciate if someone posts a better & short solution. Thank you \text{Please feel free to comment \& correct me if I am wrong \& would really appreciate if someone posts a better \& short solution. Thank you} .

Vineet PaHurKar
Jun 13, 2016

Use graph to under stand the solution of this problem,make a velocity vs time graph,,,and draw the graph of this problem v=5-2t And are under the line of eqn. at t=2to t=3 So finally ur and is o.5.....

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