141261
The \(x\)-t plot shown in the figure below describes the motion of the particle, along \(x\) axis, between two positions \(A\) and \(B\). The particle passes through two intermediate points \(P_{1}\) and \(P_{2}\) as shown in the figure.
1 The instantaneous velocity is positive as \(P_{1}\) and negative at \(\mathrm{P}_{2}\).
2 The instantaneous velocity is negative at both \(\mathrm{P}_{2}\).
3 The instantaneous velocity is negative at \(\mathrm{P}_{1}\) and positive at \(\mathrm{P}_{2}\).
4 The instantaneous velocity is positive at both \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\).
5 The instantaneous velocity is always positive.
Explanation:
A \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\) slope of \(\mathrm{x}-\mathrm{t}\) plot \(\mathrm{v}_{\mathrm{p}_{1}}=\tan \theta_{1}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{1}}\) is positive \(\Rightarrow \mathrm{v}_{\mathrm{p}_{1}}\) is postive \(\left(\because \theta_{1} \lt 90^{\circ}, \tan \theta_{1}=+\mathrm{ve}\right)\) \(\mathrm{v}_{\mathrm{p}_{2}}=\tan \theta_{2}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{2}}\) is negative \(\Rightarrow \mathrm{v}_{\mathrm{P}_{2}}\) is negative \(\left(\because \theta_{2}>90^{\circ}, \tan \theta_{2}=-\mathrm{ve}\right)\)
Kerala CEE - 2017
Motion in One Dimensions
141262
The following figure gives the movement of an object. Select the correct statement from the given choices.
1 The total distance travelled by the object is \(975 \mathrm{~m}\)
2 The maximum acceleration of the object is \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 The maximum deceleration happened between 25 th and 35 th seconds
4 The object was at rest between 10th and 15th seconds
5 At 40th second, the speed of object was decelerating
Explanation:
A Total distance travelled \(=\) Area under \(\mathrm{s}-\mathrm{t}\) graph \(\therefore\) Total distance travelled \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\mathrm{A}_{3}+\mathrm{A}_{4}+\mathrm{A}_{5}+\mathrm{A}_{6}+\mathrm{A}_{7}+\mathrm{A}_{8}\) \(=\frac{1}{2} \times 10 \times 15+(15-0)(15-10)+(15-0)(25-15)\) \(+\frac{1}{2} \times(25-15)(40-15)+(20-0)(35-25)\) \(+\frac{1}{2} \times(40-20)(35-25)+(20-0)(45-35)\) \(+\frac{1}{2}(20-0) \times(50-45)\) \(= 75+75+150+125+200+100+200+50=975 \mathrm{~m}\)
Kerala CEE - 2017
Motion in One Dimensions
141263
Consider a point \(P\), the contact point of a wheel of radius \(r\) on the ground which rolls on the ground without slipping. What is the displacement of point \(P\), when the wheel completes half rotation?
1 \(2 \mathrm{r}\)
2 \(\pi \mathrm{r}\)
3 \(\mathrm{r} \sqrt{\pi^{2}+4}\)
4 \(\mathrm{r} \sqrt{\pi^{2}+2}\)
Explanation:
C When the wheel rolls on the ground without slipping and completes half rotation, point \(P\) takes new position as \(\mathrm{P}^{\prime}\) as shown in figure. Horizontal displacement \(=\pi \mathrm{r}\) Vertical displacement \(=2 \mathrm{r}\) Thus, displacement of the point \(\mathrm{P}\) when wheel complete half rotation. \(S=\sqrt{(\pi r)^{2}+(2 r)^{2}}\) \(S=\sqrt{\pi^{2} r^{2}+4 r^{2}}\) \(S=r \sqrt{\pi^{2}+4}\)
SCRA-2012
Motion in One Dimensions
141264
The displacement time graph of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(\mathrm{X}\)-axis. The ratio of their velocities is :
1 \(\sqrt{3}: 2\)
2 \(1: 1\)
3 \(1: 2\)
4 \(1: \sqrt{3}\)
Explanation:
D By the definition, the slope of displacement time graph is velocity. i.e. \(\quad v=\frac{d x}{d t}=\tan \theta\) So, \(\quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1 / \sqrt{3}}{1}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{3}}\)
141261
The \(x\)-t plot shown in the figure below describes the motion of the particle, along \(x\) axis, between two positions \(A\) and \(B\). The particle passes through two intermediate points \(P_{1}\) and \(P_{2}\) as shown in the figure.
1 The instantaneous velocity is positive as \(P_{1}\) and negative at \(\mathrm{P}_{2}\).
2 The instantaneous velocity is negative at both \(\mathrm{P}_{2}\).
3 The instantaneous velocity is negative at \(\mathrm{P}_{1}\) and positive at \(\mathrm{P}_{2}\).
4 The instantaneous velocity is positive at both \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\).
5 The instantaneous velocity is always positive.
Explanation:
A \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\) slope of \(\mathrm{x}-\mathrm{t}\) plot \(\mathrm{v}_{\mathrm{p}_{1}}=\tan \theta_{1}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{1}}\) is positive \(\Rightarrow \mathrm{v}_{\mathrm{p}_{1}}\) is postive \(\left(\because \theta_{1} \lt 90^{\circ}, \tan \theta_{1}=+\mathrm{ve}\right)\) \(\mathrm{v}_{\mathrm{p}_{2}}=\tan \theta_{2}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{2}}\) is negative \(\Rightarrow \mathrm{v}_{\mathrm{P}_{2}}\) is negative \(\left(\because \theta_{2}>90^{\circ}, \tan \theta_{2}=-\mathrm{ve}\right)\)
Kerala CEE - 2017
Motion in One Dimensions
141262
The following figure gives the movement of an object. Select the correct statement from the given choices.
1 The total distance travelled by the object is \(975 \mathrm{~m}\)
2 The maximum acceleration of the object is \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 The maximum deceleration happened between 25 th and 35 th seconds
4 The object was at rest between 10th and 15th seconds
5 At 40th second, the speed of object was decelerating
Explanation:
A Total distance travelled \(=\) Area under \(\mathrm{s}-\mathrm{t}\) graph \(\therefore\) Total distance travelled \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\mathrm{A}_{3}+\mathrm{A}_{4}+\mathrm{A}_{5}+\mathrm{A}_{6}+\mathrm{A}_{7}+\mathrm{A}_{8}\) \(=\frac{1}{2} \times 10 \times 15+(15-0)(15-10)+(15-0)(25-15)\) \(+\frac{1}{2} \times(25-15)(40-15)+(20-0)(35-25)\) \(+\frac{1}{2} \times(40-20)(35-25)+(20-0)(45-35)\) \(+\frac{1}{2}(20-0) \times(50-45)\) \(= 75+75+150+125+200+100+200+50=975 \mathrm{~m}\)
Kerala CEE - 2017
Motion in One Dimensions
141263
Consider a point \(P\), the contact point of a wheel of radius \(r\) on the ground which rolls on the ground without slipping. What is the displacement of point \(P\), when the wheel completes half rotation?
1 \(2 \mathrm{r}\)
2 \(\pi \mathrm{r}\)
3 \(\mathrm{r} \sqrt{\pi^{2}+4}\)
4 \(\mathrm{r} \sqrt{\pi^{2}+2}\)
Explanation:
C When the wheel rolls on the ground without slipping and completes half rotation, point \(P\) takes new position as \(\mathrm{P}^{\prime}\) as shown in figure. Horizontal displacement \(=\pi \mathrm{r}\) Vertical displacement \(=2 \mathrm{r}\) Thus, displacement of the point \(\mathrm{P}\) when wheel complete half rotation. \(S=\sqrt{(\pi r)^{2}+(2 r)^{2}}\) \(S=\sqrt{\pi^{2} r^{2}+4 r^{2}}\) \(S=r \sqrt{\pi^{2}+4}\)
SCRA-2012
Motion in One Dimensions
141264
The displacement time graph of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(\mathrm{X}\)-axis. The ratio of their velocities is :
1 \(\sqrt{3}: 2\)
2 \(1: 1\)
3 \(1: 2\)
4 \(1: \sqrt{3}\)
Explanation:
D By the definition, the slope of displacement time graph is velocity. i.e. \(\quad v=\frac{d x}{d t}=\tan \theta\) So, \(\quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1 / \sqrt{3}}{1}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{3}}\)
141261
The \(x\)-t plot shown in the figure below describes the motion of the particle, along \(x\) axis, between two positions \(A\) and \(B\). The particle passes through two intermediate points \(P_{1}\) and \(P_{2}\) as shown in the figure.
1 The instantaneous velocity is positive as \(P_{1}\) and negative at \(\mathrm{P}_{2}\).
2 The instantaneous velocity is negative at both \(\mathrm{P}_{2}\).
3 The instantaneous velocity is negative at \(\mathrm{P}_{1}\) and positive at \(\mathrm{P}_{2}\).
4 The instantaneous velocity is positive at both \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\).
5 The instantaneous velocity is always positive.
Explanation:
A \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\) slope of \(\mathrm{x}-\mathrm{t}\) plot \(\mathrm{v}_{\mathrm{p}_{1}}=\tan \theta_{1}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{1}}\) is positive \(\Rightarrow \mathrm{v}_{\mathrm{p}_{1}}\) is postive \(\left(\because \theta_{1} \lt 90^{\circ}, \tan \theta_{1}=+\mathrm{ve}\right)\) \(\mathrm{v}_{\mathrm{p}_{2}}=\tan \theta_{2}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{2}}\) is negative \(\Rightarrow \mathrm{v}_{\mathrm{P}_{2}}\) is negative \(\left(\because \theta_{2}>90^{\circ}, \tan \theta_{2}=-\mathrm{ve}\right)\)
Kerala CEE - 2017
Motion in One Dimensions
141262
The following figure gives the movement of an object. Select the correct statement from the given choices.
1 The total distance travelled by the object is \(975 \mathrm{~m}\)
2 The maximum acceleration of the object is \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 The maximum deceleration happened between 25 th and 35 th seconds
4 The object was at rest between 10th and 15th seconds
5 At 40th second, the speed of object was decelerating
Explanation:
A Total distance travelled \(=\) Area under \(\mathrm{s}-\mathrm{t}\) graph \(\therefore\) Total distance travelled \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\mathrm{A}_{3}+\mathrm{A}_{4}+\mathrm{A}_{5}+\mathrm{A}_{6}+\mathrm{A}_{7}+\mathrm{A}_{8}\) \(=\frac{1}{2} \times 10 \times 15+(15-0)(15-10)+(15-0)(25-15)\) \(+\frac{1}{2} \times(25-15)(40-15)+(20-0)(35-25)\) \(+\frac{1}{2} \times(40-20)(35-25)+(20-0)(45-35)\) \(+\frac{1}{2}(20-0) \times(50-45)\) \(= 75+75+150+125+200+100+200+50=975 \mathrm{~m}\)
Kerala CEE - 2017
Motion in One Dimensions
141263
Consider a point \(P\), the contact point of a wheel of radius \(r\) on the ground which rolls on the ground without slipping. What is the displacement of point \(P\), when the wheel completes half rotation?
1 \(2 \mathrm{r}\)
2 \(\pi \mathrm{r}\)
3 \(\mathrm{r} \sqrt{\pi^{2}+4}\)
4 \(\mathrm{r} \sqrt{\pi^{2}+2}\)
Explanation:
C When the wheel rolls on the ground without slipping and completes half rotation, point \(P\) takes new position as \(\mathrm{P}^{\prime}\) as shown in figure. Horizontal displacement \(=\pi \mathrm{r}\) Vertical displacement \(=2 \mathrm{r}\) Thus, displacement of the point \(\mathrm{P}\) when wheel complete half rotation. \(S=\sqrt{(\pi r)^{2}+(2 r)^{2}}\) \(S=\sqrt{\pi^{2} r^{2}+4 r^{2}}\) \(S=r \sqrt{\pi^{2}+4}\)
SCRA-2012
Motion in One Dimensions
141264
The displacement time graph of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(\mathrm{X}\)-axis. The ratio of their velocities is :
1 \(\sqrt{3}: 2\)
2 \(1: 1\)
3 \(1: 2\)
4 \(1: \sqrt{3}\)
Explanation:
D By the definition, the slope of displacement time graph is velocity. i.e. \(\quad v=\frac{d x}{d t}=\tan \theta\) So, \(\quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1 / \sqrt{3}}{1}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{3}}\)
141261
The \(x\)-t plot shown in the figure below describes the motion of the particle, along \(x\) axis, between two positions \(A\) and \(B\). The particle passes through two intermediate points \(P_{1}\) and \(P_{2}\) as shown in the figure.
1 The instantaneous velocity is positive as \(P_{1}\) and negative at \(\mathrm{P}_{2}\).
2 The instantaneous velocity is negative at both \(\mathrm{P}_{2}\).
3 The instantaneous velocity is negative at \(\mathrm{P}_{1}\) and positive at \(\mathrm{P}_{2}\).
4 The instantaneous velocity is positive at both \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\).
5 The instantaneous velocity is always positive.
Explanation:
A \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\) slope of \(\mathrm{x}-\mathrm{t}\) plot \(\mathrm{v}_{\mathrm{p}_{1}}=\tan \theta_{1}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{1}}\) is positive \(\Rightarrow \mathrm{v}_{\mathrm{p}_{1}}\) is postive \(\left(\because \theta_{1} \lt 90^{\circ}, \tan \theta_{1}=+\mathrm{ve}\right)\) \(\mathrm{v}_{\mathrm{p}_{2}}=\tan \theta_{2}=\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)_{\mathrm{P}_{2}}\) is negative \(\Rightarrow \mathrm{v}_{\mathrm{P}_{2}}\) is negative \(\left(\because \theta_{2}>90^{\circ}, \tan \theta_{2}=-\mathrm{ve}\right)\)
Kerala CEE - 2017
Motion in One Dimensions
141262
The following figure gives the movement of an object. Select the correct statement from the given choices.
1 The total distance travelled by the object is \(975 \mathrm{~m}\)
2 The maximum acceleration of the object is \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 The maximum deceleration happened between 25 th and 35 th seconds
4 The object was at rest between 10th and 15th seconds
5 At 40th second, the speed of object was decelerating
Explanation:
A Total distance travelled \(=\) Area under \(\mathrm{s}-\mathrm{t}\) graph \(\therefore\) Total distance travelled \(=\mathrm{A}_{1}+\mathrm{A}_{2}+\mathrm{A}_{3}+\mathrm{A}_{4}+\mathrm{A}_{5}+\mathrm{A}_{6}+\mathrm{A}_{7}+\mathrm{A}_{8}\) \(=\frac{1}{2} \times 10 \times 15+(15-0)(15-10)+(15-0)(25-15)\) \(+\frac{1}{2} \times(25-15)(40-15)+(20-0)(35-25)\) \(+\frac{1}{2} \times(40-20)(35-25)+(20-0)(45-35)\) \(+\frac{1}{2}(20-0) \times(50-45)\) \(= 75+75+150+125+200+100+200+50=975 \mathrm{~m}\)
Kerala CEE - 2017
Motion in One Dimensions
141263
Consider a point \(P\), the contact point of a wheel of radius \(r\) on the ground which rolls on the ground without slipping. What is the displacement of point \(P\), when the wheel completes half rotation?
1 \(2 \mathrm{r}\)
2 \(\pi \mathrm{r}\)
3 \(\mathrm{r} \sqrt{\pi^{2}+4}\)
4 \(\mathrm{r} \sqrt{\pi^{2}+2}\)
Explanation:
C When the wheel rolls on the ground without slipping and completes half rotation, point \(P\) takes new position as \(\mathrm{P}^{\prime}\) as shown in figure. Horizontal displacement \(=\pi \mathrm{r}\) Vertical displacement \(=2 \mathrm{r}\) Thus, displacement of the point \(\mathrm{P}\) when wheel complete half rotation. \(S=\sqrt{(\pi r)^{2}+(2 r)^{2}}\) \(S=\sqrt{\pi^{2} r^{2}+4 r^{2}}\) \(S=r \sqrt{\pi^{2}+4}\)
SCRA-2012
Motion in One Dimensions
141264
The displacement time graph of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(\mathrm{X}\)-axis. The ratio of their velocities is :
1 \(\sqrt{3}: 2\)
2 \(1: 1\)
3 \(1: 2\)
4 \(1: \sqrt{3}\)
Explanation:
D By the definition, the slope of displacement time graph is velocity. i.e. \(\quad v=\frac{d x}{d t}=\tan \theta\) So, \(\quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1 / \sqrt{3}}{1}\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{3}}\)
