<p><strong>If t = <math><mrow><mroot><mrow><mi>x</mi></mrow><mrow></mrow></mroot><mo>+</mo><mn>4</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>t</mi><mi>h</mi><mi>e</mi><mi>n</mi><msub><mrow><mrow><mo>(</mo><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac></mrow><mo>)</mo></mrow></mrow><mrow><mi>t</mi><mo>=</mo><mn>4</mn></mrow></msub><mi>i</mi><mi>s</mi><mo>:</mo></mrow></math></strong></p>

t=x+4then dtdx=12x∴ (dxdt)t=4= (2x)t=4= [2 (t−4)]at t=4= 0

Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Class 12th Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Three

If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$

Option 1 -

Option 2 -

zero

Option 3 -

Option 4 -

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 2

Detailed Solution:

$t = \sqrt[]{x} + 4$

then $\frac{d t}{d x} = \frac{1}{2 \sqrt[]{x}} ∴ {(\frac{d x}{d t})}_{t = 4} = {(2 \sqrt[]{x})}_{t = 4} = {[2 (t - 4)]}_{a t t = 4}$

= 0

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[use g = 10 ms^-²] :

alok kumar singh

Let 't' be the time taken by B to meat A.

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Now for B

h = ut + $\frac{1}{2} a t^{2}$

...more

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$? 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

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less

Payal Gupta

Let ‘t’ be the time taken by B to meat

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 × t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow 80 = u \times 2 + \frac{1}{2} \times 10 t^{2}$

80 = 2u + 5 × 2 × 2

80 = 2u + 20

2u = 60

u = 30m/sec

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Height above tower

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If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$