NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI04:MOTION IN A PLANE
361818
A particle moves along the parabolic path \(y = a{x^2}\) in such a way that the \(x\)-component of the velocity remains constant, say \(c\). The acceleration of the particle is
361819
Position of an ant (in metres) moving in \(Y-Z\) plane is given by \(S=2 t^{2} \hat{j}+5 \hat{k}\) (where \(t\) is in second). The magnitude and direction of velocity of the ant at \(t=1 s\) will be :
1 \(4\;m{\rm{/}}s\) in \(x\)-direction
2 \(16\;m{\rm{/}}s\) in y-direction
3 \(9\;m{\rm{/}}s\) in z-direction
4 \(4\;m{\rm{/}}s\) in \(y\)-direction
Explanation:
Instantaneous velocity of a body is rate of change of position vector \(\vec v = \frac{{d\vec s}}{{dt}}\) \(\vec s = 2{t^2}\hat j + 5\hat k\) \(\vec v = \frac{{d\vec s}}{{dt}} = 4\hat j\) at \(t=1\) second \(\vec{v}=4 \hat{j}\)
JEE - 2024
PHXI04:MOTION IN A PLANE
361820
The co-ordinates of a particle moving in \(x-y\) plane are given by \(x=2+4 t, y=3 t+8 t^{2}\). The motion of the particle is
1 uniformly accelerated having motion along a parabolic path.
2 uniformly accelerated having motion along a straight line.
3 uniform motion along a straight line.
4 non-uniformly accelerated.
Explanation:
\(x=2+4 t, y=3 t+8 t^{2}\) Velocity and acceleration along \(x\) by axis are as follow \(v_{x}=\dfrac{d x}{d t}=4, \quad v_{y}=\dfrac{d y}{d t}=3+16 t\) \(a_{x}=\dfrac{d v_{x}}{d t}=0, \quad a_{y}=\dfrac{d v_{y}}{d t}=16\) So, \(a_{\text {net }}=16 \quad\left(\because \sqrt{a_{x}^{2}+a_{y}^{2}}\right)\) So, particle has uniformly accelerated motion. Using, \(x=2+4\) \(\Rightarrow t=\dfrac{x-2}{4}\) Putting this value in \(y\), we get \(\Rightarrow y=3\left(\dfrac{x-2}{4}\right)+8\left(\dfrac{x-2}{4}\right)^{2}\) So, path is parabolic
JEE - 2024
PHXI04:MOTION IN A PLANE
361821
A particle moves in the \(x\)-\(y\) plane with only \(x\) component of acceleration at \(2\,m/{s^2}.\) The particle starts from the origin. With an initial velocity having \(x\)-component of \(8\,m/s\) and \(y\)-component of \( - 15\,m/s\) velocity at t is
361818
A particle moves along the parabolic path \(y = a{x^2}\) in such a way that the \(x\)-component of the velocity remains constant, say \(c\). The acceleration of the particle is
361819
Position of an ant (in metres) moving in \(Y-Z\) plane is given by \(S=2 t^{2} \hat{j}+5 \hat{k}\) (where \(t\) is in second). The magnitude and direction of velocity of the ant at \(t=1 s\) will be :
1 \(4\;m{\rm{/}}s\) in \(x\)-direction
2 \(16\;m{\rm{/}}s\) in y-direction
3 \(9\;m{\rm{/}}s\) in z-direction
4 \(4\;m{\rm{/}}s\) in \(y\)-direction
Explanation:
Instantaneous velocity of a body is rate of change of position vector \(\vec v = \frac{{d\vec s}}{{dt}}\) \(\vec s = 2{t^2}\hat j + 5\hat k\) \(\vec v = \frac{{d\vec s}}{{dt}} = 4\hat j\) at \(t=1\) second \(\vec{v}=4 \hat{j}\)
JEE - 2024
PHXI04:MOTION IN A PLANE
361820
The co-ordinates of a particle moving in \(x-y\) plane are given by \(x=2+4 t, y=3 t+8 t^{2}\). The motion of the particle is
1 uniformly accelerated having motion along a parabolic path.
2 uniformly accelerated having motion along a straight line.
3 uniform motion along a straight line.
4 non-uniformly accelerated.
Explanation:
\(x=2+4 t, y=3 t+8 t^{2}\) Velocity and acceleration along \(x\) by axis are as follow \(v_{x}=\dfrac{d x}{d t}=4, \quad v_{y}=\dfrac{d y}{d t}=3+16 t\) \(a_{x}=\dfrac{d v_{x}}{d t}=0, \quad a_{y}=\dfrac{d v_{y}}{d t}=16\) So, \(a_{\text {net }}=16 \quad\left(\because \sqrt{a_{x}^{2}+a_{y}^{2}}\right)\) So, particle has uniformly accelerated motion. Using, \(x=2+4\) \(\Rightarrow t=\dfrac{x-2}{4}\) Putting this value in \(y\), we get \(\Rightarrow y=3\left(\dfrac{x-2}{4}\right)+8\left(\dfrac{x-2}{4}\right)^{2}\) So, path is parabolic
JEE - 2024
PHXI04:MOTION IN A PLANE
361821
A particle moves in the \(x\)-\(y\) plane with only \(x\) component of acceleration at \(2\,m/{s^2}.\) The particle starts from the origin. With an initial velocity having \(x\)-component of \(8\,m/s\) and \(y\)-component of \( - 15\,m/s\) velocity at t is
361818
A particle moves along the parabolic path \(y = a{x^2}\) in such a way that the \(x\)-component of the velocity remains constant, say \(c\). The acceleration of the particle is
361819
Position of an ant (in metres) moving in \(Y-Z\) plane is given by \(S=2 t^{2} \hat{j}+5 \hat{k}\) (where \(t\) is in second). The magnitude and direction of velocity of the ant at \(t=1 s\) will be :
1 \(4\;m{\rm{/}}s\) in \(x\)-direction
2 \(16\;m{\rm{/}}s\) in y-direction
3 \(9\;m{\rm{/}}s\) in z-direction
4 \(4\;m{\rm{/}}s\) in \(y\)-direction
Explanation:
Instantaneous velocity of a body is rate of change of position vector \(\vec v = \frac{{d\vec s}}{{dt}}\) \(\vec s = 2{t^2}\hat j + 5\hat k\) \(\vec v = \frac{{d\vec s}}{{dt}} = 4\hat j\) at \(t=1\) second \(\vec{v}=4 \hat{j}\)
JEE - 2024
PHXI04:MOTION IN A PLANE
361820
The co-ordinates of a particle moving in \(x-y\) plane are given by \(x=2+4 t, y=3 t+8 t^{2}\). The motion of the particle is
1 uniformly accelerated having motion along a parabolic path.
2 uniformly accelerated having motion along a straight line.
3 uniform motion along a straight line.
4 non-uniformly accelerated.
Explanation:
\(x=2+4 t, y=3 t+8 t^{2}\) Velocity and acceleration along \(x\) by axis are as follow \(v_{x}=\dfrac{d x}{d t}=4, \quad v_{y}=\dfrac{d y}{d t}=3+16 t\) \(a_{x}=\dfrac{d v_{x}}{d t}=0, \quad a_{y}=\dfrac{d v_{y}}{d t}=16\) So, \(a_{\text {net }}=16 \quad\left(\because \sqrt{a_{x}^{2}+a_{y}^{2}}\right)\) So, particle has uniformly accelerated motion. Using, \(x=2+4\) \(\Rightarrow t=\dfrac{x-2}{4}\) Putting this value in \(y\), we get \(\Rightarrow y=3\left(\dfrac{x-2}{4}\right)+8\left(\dfrac{x-2}{4}\right)^{2}\) So, path is parabolic
JEE - 2024
PHXI04:MOTION IN A PLANE
361821
A particle moves in the \(x\)-\(y\) plane with only \(x\) component of acceleration at \(2\,m/{s^2}.\) The particle starts from the origin. With an initial velocity having \(x\)-component of \(8\,m/s\) and \(y\)-component of \( - 15\,m/s\) velocity at t is
361818
A particle moves along the parabolic path \(y = a{x^2}\) in such a way that the \(x\)-component of the velocity remains constant, say \(c\). The acceleration of the particle is
361819
Position of an ant (in metres) moving in \(Y-Z\) plane is given by \(S=2 t^{2} \hat{j}+5 \hat{k}\) (where \(t\) is in second). The magnitude and direction of velocity of the ant at \(t=1 s\) will be :
1 \(4\;m{\rm{/}}s\) in \(x\)-direction
2 \(16\;m{\rm{/}}s\) in y-direction
3 \(9\;m{\rm{/}}s\) in z-direction
4 \(4\;m{\rm{/}}s\) in \(y\)-direction
Explanation:
Instantaneous velocity of a body is rate of change of position vector \(\vec v = \frac{{d\vec s}}{{dt}}\) \(\vec s = 2{t^2}\hat j + 5\hat k\) \(\vec v = \frac{{d\vec s}}{{dt}} = 4\hat j\) at \(t=1\) second \(\vec{v}=4 \hat{j}\)
JEE - 2024
PHXI04:MOTION IN A PLANE
361820
The co-ordinates of a particle moving in \(x-y\) plane are given by \(x=2+4 t, y=3 t+8 t^{2}\). The motion of the particle is
1 uniformly accelerated having motion along a parabolic path.
2 uniformly accelerated having motion along a straight line.
3 uniform motion along a straight line.
4 non-uniformly accelerated.
Explanation:
\(x=2+4 t, y=3 t+8 t^{2}\) Velocity and acceleration along \(x\) by axis are as follow \(v_{x}=\dfrac{d x}{d t}=4, \quad v_{y}=\dfrac{d y}{d t}=3+16 t\) \(a_{x}=\dfrac{d v_{x}}{d t}=0, \quad a_{y}=\dfrac{d v_{y}}{d t}=16\) So, \(a_{\text {net }}=16 \quad\left(\because \sqrt{a_{x}^{2}+a_{y}^{2}}\right)\) So, particle has uniformly accelerated motion. Using, \(x=2+4\) \(\Rightarrow t=\dfrac{x-2}{4}\) Putting this value in \(y\), we get \(\Rightarrow y=3\left(\dfrac{x-2}{4}\right)+8\left(\dfrac{x-2}{4}\right)^{2}\) So, path is parabolic
JEE - 2024
PHXI04:MOTION IN A PLANE
361821
A particle moves in the \(x\)-\(y\) plane with only \(x\) component of acceleration at \(2\,m/{s^2}.\) The particle starts from the origin. With an initial velocity having \(x\)-component of \(8\,m/s\) and \(y\)-component of \( - 15\,m/s\) velocity at t is
