At time t = 0 a particle starts travelling from a height <math><mrow><mn>7</mn><mover accent="true"><mi>z</mi><mo>^</mo></mover><mi>c</mi><mi>m</mi></mrow></math> in a plane keeping z coordinate constant. At any instant of time it’s position along the <math><mrow><mover accent="true"><mi>x</mi><mo>^</mo></mover><mtext> </mtext><mtext> </mtext><mi>a</mi><mi>n</mi><mi>d</mi><mtext> </mtext><mtext> </mtext><mover accent="true"><mi>y</mi><mo>^</mo></mover></mrow></math> directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

x = 3t y = 5t3Vx=dxdt=3Vy=dydt=15t2ax=dVxdt=0ay=dVydt=30ta→=30tj^at t = 1 sec, a→=30j^

At time t = 0 a particle starts travelling from a height&nbsp;<math><mrow><mn>7</mn><mover accent="true"><mi>z</mi><mo>^</mo></mover><mi>c</mi><mi>m</mi></mrow></math>&nbsp;in a plane keeping z coordinate constant. At any instant of time it’s position along the&nbsp;<math><mrow><mover accent="true"><mi>x</mi><mo>^</mo></mover><mtext> </mtext><mtext> </mtext><mi>a</mi><mi>n</mi><mi>d</mi><mtext> </mtext><mtext> </mtext><mover accent="true"><mi>y</mi><mo>^</mo></mover></mrow></math>&nbsp;directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

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Physics Class 12th Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

At time t = 0 a particle starts travelling from a height $7 \hat{z} c m$ in a plane keeping z coordinate constant. At any instant of time it’s position along the $\hat{x} a n d \hat{y}$ directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

Option 1 -

$- 30 \hat{y}$

Option 2 -

$30 \hat{y}$

Option 3 -

$3 \hat{x} + 15 \hat{y}$

Option 4 -

$3 \hat{x} + 15 \hat{y} + 7 \hat{z}$

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

x = 3t y = 5t³

$V_{x} = \frac{d x}{d t} = 3 V_{y} = \frac{d y}{d t} = 15 t^{2}$

$a_{x} = \frac{d V_{x}}{d t} = 0 a_{y} = \frac{d V_{y}}{d t} = 30 t$

$\vec{a} = 30 t \hat{j}$

at t = 1 sec, $\vec{a} = 30 \hat{j}$

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At time t = 0 a particle starts travelling from a height $7 \hat{z} c m$ in a plane keeping z coordinate constant. At any instant of time it’s position along the $\hat{x} a n d \hat{y}$ directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

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At time t = 0 a particle starts travelling from a height 7z^cm in a plane keeping z coordinate constant. At any instant of time it’s position along the x^ and y^ directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

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At time t = 0 a particle starts travelling from a height $7 \hat{z} c m$ in a plane keeping z coordinate constant. At any instant of time it’s position along the $\hat{x} a n d \hat{y}$ directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be