A particle is moving along x -axis with a constant acceleration of − 2 ms − 2 . The particle leaves the origin with a velocity of 5 ms − 1 . Determine the distance (in m ) travelled by the particle during the third second of its motion.
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Thank you :-) Nice solution (+1)! You loved it but you didn't liked it :(
Nice approach! (+1)!
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Thanks :):)
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Did you appeared for jee this year or you are in 11th ?
Let's see when the particle comes to rest ,
From v = u + at ,
0 = 5 − 2 t
⇒ t = 2 . 5 s ,
Hence, the particle changes its direction of motion at t = 2 . 5 s , so we cannot directly use the formula of distance travelled in n t h sec. Therefore we will calculate distance travelled from t = 2 to t = 2 . 5 & from t = 2 . 5 to t = 3 separately !
Now , we use x = u t + 2 1 a t 2
For displacement in t = 2 to t = 2 . 5 , we subtract the displacement at t = 2 from displacement at t = 2 . 5 :-
x 1 = 5 ⋅ ( 2 ) − 4 ⇒ x 1 = 6 m ....(disp. in t = 2s)
x 2 = 5 ⋅ ( 2 . 5 ) − 6 . 2 5 ⇒ x 2 = 6 . 2 5 m ....(disp. in t = 2.5s)
Required displacement = x 2 − x 1 = 0 . 2 5 m .......................... ( 1 )
Now our half work is done, For t = 2 . 5 to t = 3 , we have :
u = 0
a = + 2
t = 0 . 5
Putting these values in
x = u t + 2 1 a t 2
We get x = 0 . 2 5 m ...................... ( 2 )
Adding ( 1 ) & ( 2 ) we get net displacement in 3rd second as x = 0 . 5 m .
Please feel free to comment & correct me if I am wrong & would really appreciate if someone posts a better & short solution. Thank you .
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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Notice that the particle have initial velocity towards + x axis and acceleration towards − x axis
If we apply the direct formula to calculate the Displacement in n t h second , we get
S n = u + 2 a ( 2 n − 1 ) ⟹ S 3 = 5 − 2 2 ⋅ ( 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
v = u + a t ⟹ 0 = 5 − 2 ⋅ t t = 2 . 5 Seconds
Total Distance traveled by the particle ( S ) will be
S = 2 ( S 2 . 5 − S 2 ) S = 2 ( 5 ⋅ 2 . 5 − 2 1 ⋅ 2 ⋅ 6 . 2 5 − 5 ⋅ 2 + 2 1 ⋅ 2 ⋅ 4 ) S = 2 ( 1 2 . 5 − 6 . 2 5 − 1 0 + 4 ) S = 0 . 5 m
Nice problem , i loved it :)