141319
A bullet is fired with a velocity \(u\) making an angle of \(60^{\circ}\) with horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is
1 \(\mathrm{u}\)
2 0
3 \(\frac{\sqrt{3} u}{2}\)
4 \(u / 2\)
Explanation:
D We know that the horizontal velocity remains the same in case if the projectile motion. So, in the given case the horizontal velocity will be the horizontal component of the velocity of the Bullet. Now, the horizontal component \(=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}\) This velocity will remain same throughout the projectile motion whether it is at the maximum height or whether it is in minimum height.
WB JEE 2009
Motion in One Dimensions
141338
A particle is moving in a straight line along \(x\) axis, its position is given by \(x=2 t^{2}+2 t+4\) where, \(x\) is in metre and \(t\) in second, the acceleration of the particle is
1 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(6 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given, \(x=2 t^{2}+2 t+4\) Differentiation w.r.to ' \(t\) ' both side, \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}+2+0=4 \mathrm{t}+2\) Again Differentiation w.r.t time, \(\mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4 \mathrm{~m} / \mathrm{sec}^{2}\)
UP CPMT-2001
Motion in One Dimensions
141345
Which of the following four statements is false?
1 A body can have zero velocity and still be accelerated
2 A body can have a constant velocity and still have a varying speed
3 A body can have a constant speed and still have a varying velocity
4 The direction of the velocity of a body can change when its acceleration is constant
Explanation:
B When a body is projected vertically upwards, at highest point velocity is zero but still accelerated downwards. In uniform circular motion a body has constant speed and still have a varying velocity. In projectile motion direction of the velocity of a body changes, when its acceleration is constant.
SRM JEE - 2012
Motion in One Dimensions
141363
A car traveling on a straight path moves with uniform velocity \(v_{2}\) for some time and with velocity \(v_{2}\) of next equal time, the average velocity is given by
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Displacement in first instant \(=\mathrm{v}_{1} \mathrm{t}\) Displacement in second instant \(=v_{2} t\) \(\therefore\) Average velocity \(=\frac{\mathrm{v}_{1} \mathrm{t}+\mathrm{v}_{2} \mathrm{t}}{\mathrm{t}+\mathrm{t}}\) Average velocity \(=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141319
A bullet is fired with a velocity \(u\) making an angle of \(60^{\circ}\) with horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is
1 \(\mathrm{u}\)
2 0
3 \(\frac{\sqrt{3} u}{2}\)
4 \(u / 2\)
Explanation:
D We know that the horizontal velocity remains the same in case if the projectile motion. So, in the given case the horizontal velocity will be the horizontal component of the velocity of the Bullet. Now, the horizontal component \(=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}\) This velocity will remain same throughout the projectile motion whether it is at the maximum height or whether it is in minimum height.
WB JEE 2009
Motion in One Dimensions
141338
A particle is moving in a straight line along \(x\) axis, its position is given by \(x=2 t^{2}+2 t+4\) where, \(x\) is in metre and \(t\) in second, the acceleration of the particle is
1 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(6 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given, \(x=2 t^{2}+2 t+4\) Differentiation w.r.to ' \(t\) ' both side, \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}+2+0=4 \mathrm{t}+2\) Again Differentiation w.r.t time, \(\mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4 \mathrm{~m} / \mathrm{sec}^{2}\)
UP CPMT-2001
Motion in One Dimensions
141345
Which of the following four statements is false?
1 A body can have zero velocity and still be accelerated
2 A body can have a constant velocity and still have a varying speed
3 A body can have a constant speed and still have a varying velocity
4 The direction of the velocity of a body can change when its acceleration is constant
Explanation:
B When a body is projected vertically upwards, at highest point velocity is zero but still accelerated downwards. In uniform circular motion a body has constant speed and still have a varying velocity. In projectile motion direction of the velocity of a body changes, when its acceleration is constant.
SRM JEE - 2012
Motion in One Dimensions
141363
A car traveling on a straight path moves with uniform velocity \(v_{2}\) for some time and with velocity \(v_{2}\) of next equal time, the average velocity is given by
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Displacement in first instant \(=\mathrm{v}_{1} \mathrm{t}\) Displacement in second instant \(=v_{2} t\) \(\therefore\) Average velocity \(=\frac{\mathrm{v}_{1} \mathrm{t}+\mathrm{v}_{2} \mathrm{t}}{\mathrm{t}+\mathrm{t}}\) Average velocity \(=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}\)
141319
A bullet is fired with a velocity \(u\) making an angle of \(60^{\circ}\) with horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is
1 \(\mathrm{u}\)
2 0
3 \(\frac{\sqrt{3} u}{2}\)
4 \(u / 2\)
Explanation:
D We know that the horizontal velocity remains the same in case if the projectile motion. So, in the given case the horizontal velocity will be the horizontal component of the velocity of the Bullet. Now, the horizontal component \(=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}\) This velocity will remain same throughout the projectile motion whether it is at the maximum height or whether it is in minimum height.
WB JEE 2009
Motion in One Dimensions
141338
A particle is moving in a straight line along \(x\) axis, its position is given by \(x=2 t^{2}+2 t+4\) where, \(x\) is in metre and \(t\) in second, the acceleration of the particle is
1 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(6 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given, \(x=2 t^{2}+2 t+4\) Differentiation w.r.to ' \(t\) ' both side, \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}+2+0=4 \mathrm{t}+2\) Again Differentiation w.r.t time, \(\mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4 \mathrm{~m} / \mathrm{sec}^{2}\)
UP CPMT-2001
Motion in One Dimensions
141345
Which of the following four statements is false?
1 A body can have zero velocity and still be accelerated
2 A body can have a constant velocity and still have a varying speed
3 A body can have a constant speed and still have a varying velocity
4 The direction of the velocity of a body can change when its acceleration is constant
Explanation:
B When a body is projected vertically upwards, at highest point velocity is zero but still accelerated downwards. In uniform circular motion a body has constant speed and still have a varying velocity. In projectile motion direction of the velocity of a body changes, when its acceleration is constant.
SRM JEE - 2012
Motion in One Dimensions
141363
A car traveling on a straight path moves with uniform velocity \(v_{2}\) for some time and with velocity \(v_{2}\) of next equal time, the average velocity is given by
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Displacement in first instant \(=\mathrm{v}_{1} \mathrm{t}\) Displacement in second instant \(=v_{2} t\) \(\therefore\) Average velocity \(=\frac{\mathrm{v}_{1} \mathrm{t}+\mathrm{v}_{2} \mathrm{t}}{\mathrm{t}+\mathrm{t}}\) Average velocity \(=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}\)
141319
A bullet is fired with a velocity \(u\) making an angle of \(60^{\circ}\) with horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is
1 \(\mathrm{u}\)
2 0
3 \(\frac{\sqrt{3} u}{2}\)
4 \(u / 2\)
Explanation:
D We know that the horizontal velocity remains the same in case if the projectile motion. So, in the given case the horizontal velocity will be the horizontal component of the velocity of the Bullet. Now, the horizontal component \(=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}\) This velocity will remain same throughout the projectile motion whether it is at the maximum height or whether it is in minimum height.
WB JEE 2009
Motion in One Dimensions
141338
A particle is moving in a straight line along \(x\) axis, its position is given by \(x=2 t^{2}+2 t+4\) where, \(x\) is in metre and \(t\) in second, the acceleration of the particle is
1 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(6 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given, \(x=2 t^{2}+2 t+4\) Differentiation w.r.to ' \(t\) ' both side, \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}+2+0=4 \mathrm{t}+2\) Again Differentiation w.r.t time, \(\mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4 \mathrm{~m} / \mathrm{sec}^{2}\)
UP CPMT-2001
Motion in One Dimensions
141345
Which of the following four statements is false?
1 A body can have zero velocity and still be accelerated
2 A body can have a constant velocity and still have a varying speed
3 A body can have a constant speed and still have a varying velocity
4 The direction of the velocity of a body can change when its acceleration is constant
Explanation:
B When a body is projected vertically upwards, at highest point velocity is zero but still accelerated downwards. In uniform circular motion a body has constant speed and still have a varying velocity. In projectile motion direction of the velocity of a body changes, when its acceleration is constant.
SRM JEE - 2012
Motion in One Dimensions
141363
A car traveling on a straight path moves with uniform velocity \(v_{2}\) for some time and with velocity \(v_{2}\) of next equal time, the average velocity is given by
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Displacement in first instant \(=\mathrm{v}_{1} \mathrm{t}\) Displacement in second instant \(=v_{2} t\) \(\therefore\) Average velocity \(=\frac{\mathrm{v}_{1} \mathrm{t}+\mathrm{v}_{2} \mathrm{t}}{\mathrm{t}+\mathrm{t}}\) Average velocity \(=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}\)
