1-D Kinematics

Classical Mechanics Level 2

A particle experiences a constant acceleration for 20 seconds after starting from rest. If it travels a distance ‘X’ in first 10 seconds and a distance ‘Y’ in the next 10 seconds, then the relation b.w ‘X’ and ‘Y’is-

X=Y/2 X=Y X=Y/4 X=Y/3

Swayam Agrawal by Swayam Agrawal , 17, India

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1 solution

Denis Kartachov
Jun 30, 2018

Can use basic kinematics equation Δ x = v 0 t + 1 2 a t 2 \Delta x = v_0 t + \frac{1}{2} a t^2

From position 0 to position X+Y in 20 seconds:

( X + Y ) − 0 = ( 0 ) ( 20 ) + 1 2 a ( 20 ) 2 = 200 a X + Y = 200 a \begin{array}{c} (X+Y)-0 = (0)(20) + \frac{1}{2}a(20)^2 = 200a \\ X + Y = 200a \\ \end{array}

From position 0 to X in 10 seconds:

X − 0 = ( 0 ) ( 10 ) + 1 2 a ( 10 ) 2 = 50 a a = X 50 \begin{array}{c} X-0 = (0)(10) + \frac{1}{2}a(10)^2 = 50a \\ a = \frac{X}{50} \\ \end{array}

Therefore:

X + Y = 200 ⋅ X 50 X + Y = 4 X Y = 3 X \begin{array}{c} X + Y = 200 \cdot \frac{X}{50} \\ X + Y = 4X \\ Y = 3X \\ \end{array}

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