141266
A body is moving along a straight line path with constant velocity. At an instant of time the distance travelled by it is \(\mathrm{S}\) and its displacement is \(D\), then
1 \(\mathrm{D} \lt \mathrm{S}\)
2 \(\mathrm{D}>\mathrm{S}\)
3 \(\mathrm{D}=\mathrm{S}\)
4 \(\mathrm{D} \leq \mathrm{S}\)
Explanation:
C The body is moving in straight line with uniform velocity. \(\mathrm{A} \stackrel{ }{\longleftarrow} \mathrm{S}=\mathrm{D} \longrightarrow \mathrm{B}\) So, at any instant of the time distance covered by the body is always equal to displacement. \(D=S\)
J and K-CET- 2008
Motion in One Dimensions
141268
When a particle returns to its initial point, its
1 Displacement is zero
2 Average velocity is zero
3 Distance is zero
4 Average speed zero
Explanation:
A When initial and final positions are same then displacement and average velocity both are zero. Average velocity \(\overrightarrow{\mathrm{v}}=\frac{\text { Total displacement }}{\text { Total time }}\) \(\vec{v}=\frac{0}{t}=0\)
J and K-CET-2014
Motion in One Dimensions
141288
For a body starting from rest, what will be the ratio of the distance travelled by the body during the 4th and 3rd second during its journey?
1 \(\frac{7}{5}\)
2 \(\frac{7}{3}\)
3 \(\frac{5}{7}\)
4 \(\frac{3}{7}\)
Explanation:
A We know that the distance travelled by a particle in nth second is given as \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)\) Thus, the ratio of distance travelled in \(4^{\text {th }}\) to \(3^{\text {rd }}\) second is \(\frac{\mathrm{S}_{4^{\text {th }}}}{\mathrm{S}_{3^{\text {rd }}}}=\frac{0+\frac{\mathrm{a}}{2}(2 \times 4-1)}{0+\frac{\mathrm{a}}{2}(2 \times 3-1)}=\frac{2 \times 4-1}{2 \times 3-1}=\frac{7}{5}\)
BCECE-2012
Motion in One Dimensions
141295
A bomb is dropped from an aircraft travelling horizontally at \(150 \mathrm{~ms}^{-1}\) at a height of \(490 \mathrm{~m}\). The horizontal distance travelled by the bomb before it hits the ground is (in metre)
141266
A body is moving along a straight line path with constant velocity. At an instant of time the distance travelled by it is \(\mathrm{S}\) and its displacement is \(D\), then
1 \(\mathrm{D} \lt \mathrm{S}\)
2 \(\mathrm{D}>\mathrm{S}\)
3 \(\mathrm{D}=\mathrm{S}\)
4 \(\mathrm{D} \leq \mathrm{S}\)
Explanation:
C The body is moving in straight line with uniform velocity. \(\mathrm{A} \stackrel{ }{\longleftarrow} \mathrm{S}=\mathrm{D} \longrightarrow \mathrm{B}\) So, at any instant of the time distance covered by the body is always equal to displacement. \(D=S\)
J and K-CET- 2008
Motion in One Dimensions
141268
When a particle returns to its initial point, its
1 Displacement is zero
2 Average velocity is zero
3 Distance is zero
4 Average speed zero
Explanation:
A When initial and final positions are same then displacement and average velocity both are zero. Average velocity \(\overrightarrow{\mathrm{v}}=\frac{\text { Total displacement }}{\text { Total time }}\) \(\vec{v}=\frac{0}{t}=0\)
J and K-CET-2014
Motion in One Dimensions
141288
For a body starting from rest, what will be the ratio of the distance travelled by the body during the 4th and 3rd second during its journey?
1 \(\frac{7}{5}\)
2 \(\frac{7}{3}\)
3 \(\frac{5}{7}\)
4 \(\frac{3}{7}\)
Explanation:
A We know that the distance travelled by a particle in nth second is given as \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)\) Thus, the ratio of distance travelled in \(4^{\text {th }}\) to \(3^{\text {rd }}\) second is \(\frac{\mathrm{S}_{4^{\text {th }}}}{\mathrm{S}_{3^{\text {rd }}}}=\frac{0+\frac{\mathrm{a}}{2}(2 \times 4-1)}{0+\frac{\mathrm{a}}{2}(2 \times 3-1)}=\frac{2 \times 4-1}{2 \times 3-1}=\frac{7}{5}\)
BCECE-2012
Motion in One Dimensions
141295
A bomb is dropped from an aircraft travelling horizontally at \(150 \mathrm{~ms}^{-1}\) at a height of \(490 \mathrm{~m}\). The horizontal distance travelled by the bomb before it hits the ground is (in metre)
141266
A body is moving along a straight line path with constant velocity. At an instant of time the distance travelled by it is \(\mathrm{S}\) and its displacement is \(D\), then
1 \(\mathrm{D} \lt \mathrm{S}\)
2 \(\mathrm{D}>\mathrm{S}\)
3 \(\mathrm{D}=\mathrm{S}\)
4 \(\mathrm{D} \leq \mathrm{S}\)
Explanation:
C The body is moving in straight line with uniform velocity. \(\mathrm{A} \stackrel{ }{\longleftarrow} \mathrm{S}=\mathrm{D} \longrightarrow \mathrm{B}\) So, at any instant of the time distance covered by the body is always equal to displacement. \(D=S\)
J and K-CET- 2008
Motion in One Dimensions
141268
When a particle returns to its initial point, its
1 Displacement is zero
2 Average velocity is zero
3 Distance is zero
4 Average speed zero
Explanation:
A When initial and final positions are same then displacement and average velocity both are zero. Average velocity \(\overrightarrow{\mathrm{v}}=\frac{\text { Total displacement }}{\text { Total time }}\) \(\vec{v}=\frac{0}{t}=0\)
J and K-CET-2014
Motion in One Dimensions
141288
For a body starting from rest, what will be the ratio of the distance travelled by the body during the 4th and 3rd second during its journey?
1 \(\frac{7}{5}\)
2 \(\frac{7}{3}\)
3 \(\frac{5}{7}\)
4 \(\frac{3}{7}\)
Explanation:
A We know that the distance travelled by a particle in nth second is given as \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)\) Thus, the ratio of distance travelled in \(4^{\text {th }}\) to \(3^{\text {rd }}\) second is \(\frac{\mathrm{S}_{4^{\text {th }}}}{\mathrm{S}_{3^{\text {rd }}}}=\frac{0+\frac{\mathrm{a}}{2}(2 \times 4-1)}{0+\frac{\mathrm{a}}{2}(2 \times 3-1)}=\frac{2 \times 4-1}{2 \times 3-1}=\frac{7}{5}\)
BCECE-2012
Motion in One Dimensions
141295
A bomb is dropped from an aircraft travelling horizontally at \(150 \mathrm{~ms}^{-1}\) at a height of \(490 \mathrm{~m}\). The horizontal distance travelled by the bomb before it hits the ground is (in metre)
141266
A body is moving along a straight line path with constant velocity. At an instant of time the distance travelled by it is \(\mathrm{S}\) and its displacement is \(D\), then
1 \(\mathrm{D} \lt \mathrm{S}\)
2 \(\mathrm{D}>\mathrm{S}\)
3 \(\mathrm{D}=\mathrm{S}\)
4 \(\mathrm{D} \leq \mathrm{S}\)
Explanation:
C The body is moving in straight line with uniform velocity. \(\mathrm{A} \stackrel{ }{\longleftarrow} \mathrm{S}=\mathrm{D} \longrightarrow \mathrm{B}\) So, at any instant of the time distance covered by the body is always equal to displacement. \(D=S\)
J and K-CET- 2008
Motion in One Dimensions
141268
When a particle returns to its initial point, its
1 Displacement is zero
2 Average velocity is zero
3 Distance is zero
4 Average speed zero
Explanation:
A When initial and final positions are same then displacement and average velocity both are zero. Average velocity \(\overrightarrow{\mathrm{v}}=\frac{\text { Total displacement }}{\text { Total time }}\) \(\vec{v}=\frac{0}{t}=0\)
J and K-CET-2014
Motion in One Dimensions
141288
For a body starting from rest, what will be the ratio of the distance travelled by the body during the 4th and 3rd second during its journey?
1 \(\frac{7}{5}\)
2 \(\frac{7}{3}\)
3 \(\frac{5}{7}\)
4 \(\frac{3}{7}\)
Explanation:
A We know that the distance travelled by a particle in nth second is given as \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)\) Thus, the ratio of distance travelled in \(4^{\text {th }}\) to \(3^{\text {rd }}\) second is \(\frac{\mathrm{S}_{4^{\text {th }}}}{\mathrm{S}_{3^{\text {rd }}}}=\frac{0+\frac{\mathrm{a}}{2}(2 \times 4-1)}{0+\frac{\mathrm{a}}{2}(2 \times 3-1)}=\frac{2 \times 4-1}{2 \times 3-1}=\frac{7}{5}\)
BCECE-2012
Motion in One Dimensions
141295
A bomb is dropped from an aircraft travelling horizontally at \(150 \mathrm{~ms}^{-1}\) at a height of \(490 \mathrm{~m}\). The horizontal distance travelled by the bomb before it hits the ground is (in metre)
