Question:
At time t = 0 a particle starts travelling from a height \(7\hat z cm\) in a plane keeping \(z\) coordinate constant. At any instant time it's position along the \(x\) and \(y\) directions are defined as \(3t\) and \(5t^3\) respectively. At \(t = 1s\) acceleration of the particle will be
At time t = 0 a particle starts travelling from a height \(7\hat z cm\) in a plane keeping \(z\) coordinate constant. At any instant time it's position along the \(x\) and \(y\) directions are defined as \(3t\) and \(5t^3\) respectively. At \(t = 1s\) acceleration of the particle will be
Updated On: Jul 8, 2024
- \(-30 \hat {y}\)
- \(30\hat {y}\)
- \(3 \hat {x}+ 15\hat {y}\)
- \(3\hat{x}^+15\hat{y}+7\hat{z}\)
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The Correct Option is B
Solution and Explanation
\(\vec r=3t\hat {i}+5t^3\hat{j}+7\hat{k}\)
\(\frac{d^2\vec r}{dt^2}=30t\hat{j}\) At \(t=1\)
⇒ \(\frac{d^2\vec r}{dt^2}\)= \(30\hat{j}\)
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration