362211
If a particle travels a linear distance at speed \({v_1}\) and comes back along the same track at speed \({v_2}\).
1 Its average speed is \(\left( {{v_{\rm{1}}} + {v_2}} \right)/2\)
2 Its average speed is \({v_{\rm{1}}}{v_2}/\left( {{v_{\rm{1}}} + {v_2}} \right)\)
3 Its average speed is \(\sqrt {{v_{\rm{1}}}{v_2}} \)
4 Its average velocity is zero
Explanation:
Let 2\(x\) is the total distance travelled by the particle Average Speed \( = \frac{{{\rm{Total \; distance}}}}{{{\rm{Total \; time}}}} = \frac{{2x}}{{\frac{x}{{{v_1}}} + \frac{x}{{{v_2}}}}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\) Average velocity\( = \frac{{{\rm{Total \; displacement}}}}{{{\rm{Total \; time}}}}\) It comes back to its initial position. Total displacement is zero. Hence, average velocity is zero.
PHXI03:MOTION IN A STRAIGHT LINE
362212
For a body moving with uniform acceleration ‘\(a\)’, initial and final velocities in a time interval ‘\(t\)’ are \(u\) and \(v\) respectively. The average velocity in the time interval ‘\(t\)’ is
1 \(u + at\)
2 \(u - \frac{{at}}{2}\)
3 \(u - at\)
4 \(u + \frac{{at}}{2}\)
Explanation:
The velocity of the body is \(v = u + at\) As the velocity function is given The average velocity is \(\left\langle v \right\rangle = \frac{{\int\limits_0^{t} vdt}}{t - 0} = \frac{1}{t}\int\limits_0^{t} {\left( {u + at} \right)dt} \) \( = \frac{1}{t} \left[ {\int\limits_0^{t} udt + a}\int\limits_0^{t} tdt \right] = u + \frac{{at}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362213
A point traversed \(3/{4^{th}}\) of the circle of radius \(R\) in time \(t\). The magnitude of the average velocity of the particle in this time interval is
362214
An object moves with speed \(v_{1}, v_{2}\) and \(v_{3}\) along a line segment \(A B, B C\) and \(C D\) respectively as shown in figure. Where \(A B=B C\) and \(A D=3 A B\) then average speed of the object will be
362211
If a particle travels a linear distance at speed \({v_1}\) and comes back along the same track at speed \({v_2}\).
1 Its average speed is \(\left( {{v_{\rm{1}}} + {v_2}} \right)/2\)
2 Its average speed is \({v_{\rm{1}}}{v_2}/\left( {{v_{\rm{1}}} + {v_2}} \right)\)
3 Its average speed is \(\sqrt {{v_{\rm{1}}}{v_2}} \)
4 Its average velocity is zero
Explanation:
Let 2\(x\) is the total distance travelled by the particle Average Speed \( = \frac{{{\rm{Total \; distance}}}}{{{\rm{Total \; time}}}} = \frac{{2x}}{{\frac{x}{{{v_1}}} + \frac{x}{{{v_2}}}}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\) Average velocity\( = \frac{{{\rm{Total \; displacement}}}}{{{\rm{Total \; time}}}}\) It comes back to its initial position. Total displacement is zero. Hence, average velocity is zero.
PHXI03:MOTION IN A STRAIGHT LINE
362212
For a body moving with uniform acceleration ‘\(a\)’, initial and final velocities in a time interval ‘\(t\)’ are \(u\) and \(v\) respectively. The average velocity in the time interval ‘\(t\)’ is
1 \(u + at\)
2 \(u - \frac{{at}}{2}\)
3 \(u - at\)
4 \(u + \frac{{at}}{2}\)
Explanation:
The velocity of the body is \(v = u + at\) As the velocity function is given The average velocity is \(\left\langle v \right\rangle = \frac{{\int\limits_0^{t} vdt}}{t - 0} = \frac{1}{t}\int\limits_0^{t} {\left( {u + at} \right)dt} \) \( = \frac{1}{t} \left[ {\int\limits_0^{t} udt + a}\int\limits_0^{t} tdt \right] = u + \frac{{at}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362213
A point traversed \(3/{4^{th}}\) of the circle of radius \(R\) in time \(t\). The magnitude of the average velocity of the particle in this time interval is
362214
An object moves with speed \(v_{1}, v_{2}\) and \(v_{3}\) along a line segment \(A B, B C\) and \(C D\) respectively as shown in figure. Where \(A B=B C\) and \(A D=3 A B\) then average speed of the object will be
362211
If a particle travels a linear distance at speed \({v_1}\) and comes back along the same track at speed \({v_2}\).
1 Its average speed is \(\left( {{v_{\rm{1}}} + {v_2}} \right)/2\)
2 Its average speed is \({v_{\rm{1}}}{v_2}/\left( {{v_{\rm{1}}} + {v_2}} \right)\)
3 Its average speed is \(\sqrt {{v_{\rm{1}}}{v_2}} \)
4 Its average velocity is zero
Explanation:
Let 2\(x\) is the total distance travelled by the particle Average Speed \( = \frac{{{\rm{Total \; distance}}}}{{{\rm{Total \; time}}}} = \frac{{2x}}{{\frac{x}{{{v_1}}} + \frac{x}{{{v_2}}}}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\) Average velocity\( = \frac{{{\rm{Total \; displacement}}}}{{{\rm{Total \; time}}}}\) It comes back to its initial position. Total displacement is zero. Hence, average velocity is zero.
PHXI03:MOTION IN A STRAIGHT LINE
362212
For a body moving with uniform acceleration ‘\(a\)’, initial and final velocities in a time interval ‘\(t\)’ are \(u\) and \(v\) respectively. The average velocity in the time interval ‘\(t\)’ is
1 \(u + at\)
2 \(u - \frac{{at}}{2}\)
3 \(u - at\)
4 \(u + \frac{{at}}{2}\)
Explanation:
The velocity of the body is \(v = u + at\) As the velocity function is given The average velocity is \(\left\langle v \right\rangle = \frac{{\int\limits_0^{t} vdt}}{t - 0} = \frac{1}{t}\int\limits_0^{t} {\left( {u + at} \right)dt} \) \( = \frac{1}{t} \left[ {\int\limits_0^{t} udt + a}\int\limits_0^{t} tdt \right] = u + \frac{{at}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362213
A point traversed \(3/{4^{th}}\) of the circle of radius \(R\) in time \(t\). The magnitude of the average velocity of the particle in this time interval is
362214
An object moves with speed \(v_{1}, v_{2}\) and \(v_{3}\) along a line segment \(A B, B C\) and \(C D\) respectively as shown in figure. Where \(A B=B C\) and \(A D=3 A B\) then average speed of the object will be
362211
If a particle travels a linear distance at speed \({v_1}\) and comes back along the same track at speed \({v_2}\).
1 Its average speed is \(\left( {{v_{\rm{1}}} + {v_2}} \right)/2\)
2 Its average speed is \({v_{\rm{1}}}{v_2}/\left( {{v_{\rm{1}}} + {v_2}} \right)\)
3 Its average speed is \(\sqrt {{v_{\rm{1}}}{v_2}} \)
4 Its average velocity is zero
Explanation:
Let 2\(x\) is the total distance travelled by the particle Average Speed \( = \frac{{{\rm{Total \; distance}}}}{{{\rm{Total \; time}}}} = \frac{{2x}}{{\frac{x}{{{v_1}}} + \frac{x}{{{v_2}}}}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\) Average velocity\( = \frac{{{\rm{Total \; displacement}}}}{{{\rm{Total \; time}}}}\) It comes back to its initial position. Total displacement is zero. Hence, average velocity is zero.
PHXI03:MOTION IN A STRAIGHT LINE
362212
For a body moving with uniform acceleration ‘\(a\)’, initial and final velocities in a time interval ‘\(t\)’ are \(u\) and \(v\) respectively. The average velocity in the time interval ‘\(t\)’ is
1 \(u + at\)
2 \(u - \frac{{at}}{2}\)
3 \(u - at\)
4 \(u + \frac{{at}}{2}\)
Explanation:
The velocity of the body is \(v = u + at\) As the velocity function is given The average velocity is \(\left\langle v \right\rangle = \frac{{\int\limits_0^{t} vdt}}{t - 0} = \frac{1}{t}\int\limits_0^{t} {\left( {u + at} \right)dt} \) \( = \frac{1}{t} \left[ {\int\limits_0^{t} udt + a}\int\limits_0^{t} tdt \right] = u + \frac{{at}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362213
A point traversed \(3/{4^{th}}\) of the circle of radius \(R\) in time \(t\). The magnitude of the average velocity of the particle in this time interval is
362214
An object moves with speed \(v_{1}, v_{2}\) and \(v_{3}\) along a line segment \(A B, B C\) and \(C D\) respectively as shown in figure. Where \(A B=B C\) and \(A D=3 A B\) then average speed of the object will be
362211
If a particle travels a linear distance at speed \({v_1}\) and comes back along the same track at speed \({v_2}\).
1 Its average speed is \(\left( {{v_{\rm{1}}} + {v_2}} \right)/2\)
2 Its average speed is \({v_{\rm{1}}}{v_2}/\left( {{v_{\rm{1}}} + {v_2}} \right)\)
3 Its average speed is \(\sqrt {{v_{\rm{1}}}{v_2}} \)
4 Its average velocity is zero
Explanation:
Let 2\(x\) is the total distance travelled by the particle Average Speed \( = \frac{{{\rm{Total \; distance}}}}{{{\rm{Total \; time}}}} = \frac{{2x}}{{\frac{x}{{{v_1}}} + \frac{x}{{{v_2}}}}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\) Average velocity\( = \frac{{{\rm{Total \; displacement}}}}{{{\rm{Total \; time}}}}\) It comes back to its initial position. Total displacement is zero. Hence, average velocity is zero.
PHXI03:MOTION IN A STRAIGHT LINE
362212
For a body moving with uniform acceleration ‘\(a\)’, initial and final velocities in a time interval ‘\(t\)’ are \(u\) and \(v\) respectively. The average velocity in the time interval ‘\(t\)’ is
1 \(u + at\)
2 \(u - \frac{{at}}{2}\)
3 \(u - at\)
4 \(u + \frac{{at}}{2}\)
Explanation:
The velocity of the body is \(v = u + at\) As the velocity function is given The average velocity is \(\left\langle v \right\rangle = \frac{{\int\limits_0^{t} vdt}}{t - 0} = \frac{1}{t}\int\limits_0^{t} {\left( {u + at} \right)dt} \) \( = \frac{1}{t} \left[ {\int\limits_0^{t} udt + a}\int\limits_0^{t} tdt \right] = u + \frac{{at}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362213
A point traversed \(3/{4^{th}}\) of the circle of radius \(R\) in time \(t\). The magnitude of the average velocity of the particle in this time interval is
362214
An object moves with speed \(v_{1}, v_{2}\) and \(v_{3}\) along a line segment \(A B, B C\) and \(C D\) respectively as shown in figure. Where \(A B=B C\) and \(A D=3 A B\) then average speed of the object will be
