A ball is spun with angular acceleration a = 6t2 – 2t where t is in second and a is in rads-1. At t = 0, the ball has angular velocity of 10 rads-1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

a = 6t2 – 2t, of t 20, w = 10 rad/sec , q = 4 rad α=dωdt ⇒∫10ωdω=∫t−0tαdt ⇒ω−10=∫0t (6t2−2t)dt ω=dθdt dθ=ω dt ∫4θdθ=∫0t (2t3−t2+10)dt = t42−t33+10t+4

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>−</mo> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mi>t</mi> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mi>t</mi> <mo>+</mo> <mn>4</mn> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mn>2</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mi>t</mi> <mo>+</mo> <mn>1</mn> <mn>2</mn> </mrow> </math>

Option 4 - <math> <mrow> <mn>2</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>−</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mn>5</mn> <mi>t</mi> <mo>+</mo> <mn>4</mn> </mrow> </math>

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Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

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