Question

Graph of v3 vs t' for a particle moving along a straight line is shown in the figure. Acceleration of the particle at t = 4 second is
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Answer 1

Given:

Graph of v³ vs t' for a particle moving along a straight line is shown in the figure.

To find:

Acceleration at t = 4 sec?

Calculation:

So, the slope at t = 4 sec (i.e. point P) will be :

$\rm \: \dfrac{d {v}^{3} }{dt} = \tan( {37}^{ \circ} )$

$\rm \implies\: \dfrac{d {v}^{3} }{dv} \times \dfrac{dv}{dt} = \dfrac{3}{4}$

$\rm \implies\: 3 {v}^{2} \times \dfrac{dv}{dt} = \dfrac{3}{4}$

At point P , v³ = ⅛ => v = ½ m/s , hence v² = ¼.

$\rm \implies\: \dfrac{3}{4} \times \dfrac{dv}{dt} = \dfrac{3}{4}$

$\rm \implies\: \dfrac{dv}{dt} = 1 \: m {s}^{ - 2}$

$\rm \implies\: acc. = 1 \: m {s}^{ - 2}$

So, acceleration is 1 m/s².