141188
\(\quad \mathbf{y}=\left(\mathbf{P t}^{2}-\mathbf{Q t}^{3}\right) \mathbf{m}\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
1 \(\frac{27 \mathrm{P}^{3}}{4 \mathrm{Q}^{2}}\)
2 \(\frac{4 Q^{2}}{27 \mathrm{P}^{3}}\)
3 \(\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
4 \(\frac{27 \mathrm{Q}^{2}}{4 \mathrm{P}^{3}}\)
Explanation:
C In vertical plane the vertical displacement of ball is given as - \(\mathrm{y}=\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)\) For maximum height, \(\frac{\mathrm{dy}}{\mathrm{dt}}=0\) \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)=0\) \(\left(2 \mathrm{Pt}-3 \mathrm{Qt}^{2}\right)=0\) \(\mathrm{t}(2 \mathrm{P}-3 \mathrm{Qt})=0\) \(2 \mathrm{P}-3 \mathrm{Qt}=0\) \(\mathrm{t}=\frac{2 \mathrm{P}}{3 \mathrm{Q}}\) Putting the value of \(t\) in equation (i), \(y_{\max } =\frac{4 \mathrm{P}^{3}}{9 \mathrm{Q}^{2}}-\frac{8 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}[3-2]\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
AP EAMCET-04.07.2022
Motion in One Dimensions
141189
\(\quad\) A person moved from \(A\) to \(B\) on a circular path as shown in figure. If the distance travelled by him is \(60 \mathrm{~m}\), then the magnitude of displacement would be: \(\left(\right.\) Given \(\left.\cos 135^{\circ}=-0.7\right)\)
1 \(42 \mathrm{~m}\)
2 \(47 \mathrm{~m}\)
3 \(19 \mathrm{~m}\)
4 \(40 \mathrm{~m}\)
Explanation:
B Distance travelled by Person \(=60 \mathrm{~m}\) (Arc ACB) Let, \(\mathrm{R}=\) Radius Displacement is shortest distance between two point. So, Displacement of person \(=\mathrm{AB}\) (Straight line) Using consines rule in triangle \(\mathrm{BOA}\). \((\mathrm{AB})^{2}=(\mathrm{OA})^{2}+(\mathrm{OB})^{2}-2(\mathrm{OA})(\mathrm{OB}) \cos (\mathrm{BOA})\) \((A B)^{2}=R^{2}+R^{2}-2 R \times R \cos 135^{\circ}\) \((A B)^{2}=2 R^{2}-2 R^{2} \times(-0.7) \quad\left(\because \cos 135^{\circ}=-0.7\right)\) \(\mathrm{AB}=1.84 \mathrm{R}\) Circumference of circle \(=2 \pi R\) \(\mathrm{R}=60 \times \frac{360}{135 \times 2 \pi}=\frac{80}{\pi}\) \(=\frac{80}{22} \times 7\) \(\mathrm{R}=25.45 \mathrm{~m}\) Putting value of \(R\) in \(\mathrm{eq}^{\mathrm{n}}\) (i) we have. \(\mathrm{AB} =1.84 \mathrm{R}=1.84 \times 25.45\) \(\mathrm{AB} =46.92 \mathrm{~m}\) \(\mathrm{AB} \cong 47 \mathrm{~m}\)
JEE Main-25.07.2022
Motion in One Dimensions
141191
The relation between time \(t\) and distance \(x\) for a moving body is given as \(t=\mathbf{m x}^{2}+\mathbf{n x}\), where \(m\) and \(n\) are constants. The retardation of the motion is (when \(v\) stands for velocity)
1 \(2 \mathrm{mv}^{3}\)
2 \(2 \mathrm{mnv}^{3}\)
3 \(2 \mathrm{nv}^{3}\)
4 \(2 n^{2} v^{3}\)
Explanation:
A Given, \(\mathrm{t}=m \mathrm{x}^{2}+\mathrm{nx}\) Differentiating the above equation with respect to time, we have \(\frac{\mathrm{dt}}{\mathrm{dt}}=\mathrm{m} \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=2 \mathrm{mx} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{mxv}+\mathrm{nv}\) \(1=\mathrm{v}(2 \mathrm{mx}+\mathrm{n})\) \(\mathrm{v}=\frac{1}{2 \mathrm{mx}+\mathrm{n}}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{-2 \mathrm{~m}(\mathrm{dx} / \mathrm{dt})}{(2 \mathrm{mx}+\mathrm{n})^{2}}=-2 \mathrm{~m} \times \mathrm{v}^{2} \times \mathrm{v}=-2 \mathrm{mv}^{3}\) \(\therefore\) The retardation of the motion is \(2 \mathrm{mv}^{3}\).
JEE Main-25.07.2021
Motion in One Dimensions
141192
The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between \(t=0 \mathrm{~s}\) and \(t=10\) s will be
1 \(120 \mathrm{~m}\)
2 \(90 \mathrm{~m}\)
3 \(60 \mathrm{~m}\)
4 \(30 \mathrm{~m}\)
Explanation:
B Distance travelled by the particle \(=\) Area under the v-t graph Area of \(\square \mathrm{ABCO}+\) area of \(\triangle \mathrm{BCP}\) \(=5 \times 12+\frac{1}{2} \times 12 \times 5\) \(=60+30=90 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141188
\(\quad \mathbf{y}=\left(\mathbf{P t}^{2}-\mathbf{Q t}^{3}\right) \mathbf{m}\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
1 \(\frac{27 \mathrm{P}^{3}}{4 \mathrm{Q}^{2}}\)
2 \(\frac{4 Q^{2}}{27 \mathrm{P}^{3}}\)
3 \(\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
4 \(\frac{27 \mathrm{Q}^{2}}{4 \mathrm{P}^{3}}\)
Explanation:
C In vertical plane the vertical displacement of ball is given as - \(\mathrm{y}=\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)\) For maximum height, \(\frac{\mathrm{dy}}{\mathrm{dt}}=0\) \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)=0\) \(\left(2 \mathrm{Pt}-3 \mathrm{Qt}^{2}\right)=0\) \(\mathrm{t}(2 \mathrm{P}-3 \mathrm{Qt})=0\) \(2 \mathrm{P}-3 \mathrm{Qt}=0\) \(\mathrm{t}=\frac{2 \mathrm{P}}{3 \mathrm{Q}}\) Putting the value of \(t\) in equation (i), \(y_{\max } =\frac{4 \mathrm{P}^{3}}{9 \mathrm{Q}^{2}}-\frac{8 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}[3-2]\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
AP EAMCET-04.07.2022
Motion in One Dimensions
141189
\(\quad\) A person moved from \(A\) to \(B\) on a circular path as shown in figure. If the distance travelled by him is \(60 \mathrm{~m}\), then the magnitude of displacement would be: \(\left(\right.\) Given \(\left.\cos 135^{\circ}=-0.7\right)\)
1 \(42 \mathrm{~m}\)
2 \(47 \mathrm{~m}\)
3 \(19 \mathrm{~m}\)
4 \(40 \mathrm{~m}\)
Explanation:
B Distance travelled by Person \(=60 \mathrm{~m}\) (Arc ACB) Let, \(\mathrm{R}=\) Radius Displacement is shortest distance between two point. So, Displacement of person \(=\mathrm{AB}\) (Straight line) Using consines rule in triangle \(\mathrm{BOA}\). \((\mathrm{AB})^{2}=(\mathrm{OA})^{2}+(\mathrm{OB})^{2}-2(\mathrm{OA})(\mathrm{OB}) \cos (\mathrm{BOA})\) \((A B)^{2}=R^{2}+R^{2}-2 R \times R \cos 135^{\circ}\) \((A B)^{2}=2 R^{2}-2 R^{2} \times(-0.7) \quad\left(\because \cos 135^{\circ}=-0.7\right)\) \(\mathrm{AB}=1.84 \mathrm{R}\) Circumference of circle \(=2 \pi R\) \(\mathrm{R}=60 \times \frac{360}{135 \times 2 \pi}=\frac{80}{\pi}\) \(=\frac{80}{22} \times 7\) \(\mathrm{R}=25.45 \mathrm{~m}\) Putting value of \(R\) in \(\mathrm{eq}^{\mathrm{n}}\) (i) we have. \(\mathrm{AB} =1.84 \mathrm{R}=1.84 \times 25.45\) \(\mathrm{AB} =46.92 \mathrm{~m}\) \(\mathrm{AB} \cong 47 \mathrm{~m}\)
JEE Main-25.07.2022
Motion in One Dimensions
141191
The relation between time \(t\) and distance \(x\) for a moving body is given as \(t=\mathbf{m x}^{2}+\mathbf{n x}\), where \(m\) and \(n\) are constants. The retardation of the motion is (when \(v\) stands for velocity)
1 \(2 \mathrm{mv}^{3}\)
2 \(2 \mathrm{mnv}^{3}\)
3 \(2 \mathrm{nv}^{3}\)
4 \(2 n^{2} v^{3}\)
Explanation:
A Given, \(\mathrm{t}=m \mathrm{x}^{2}+\mathrm{nx}\) Differentiating the above equation with respect to time, we have \(\frac{\mathrm{dt}}{\mathrm{dt}}=\mathrm{m} \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=2 \mathrm{mx} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{mxv}+\mathrm{nv}\) \(1=\mathrm{v}(2 \mathrm{mx}+\mathrm{n})\) \(\mathrm{v}=\frac{1}{2 \mathrm{mx}+\mathrm{n}}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{-2 \mathrm{~m}(\mathrm{dx} / \mathrm{dt})}{(2 \mathrm{mx}+\mathrm{n})^{2}}=-2 \mathrm{~m} \times \mathrm{v}^{2} \times \mathrm{v}=-2 \mathrm{mv}^{3}\) \(\therefore\) The retardation of the motion is \(2 \mathrm{mv}^{3}\).
JEE Main-25.07.2021
Motion in One Dimensions
141192
The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between \(t=0 \mathrm{~s}\) and \(t=10\) s will be
1 \(120 \mathrm{~m}\)
2 \(90 \mathrm{~m}\)
3 \(60 \mathrm{~m}\)
4 \(30 \mathrm{~m}\)
Explanation:
B Distance travelled by the particle \(=\) Area under the v-t graph Area of \(\square \mathrm{ABCO}+\) area of \(\triangle \mathrm{BCP}\) \(=5 \times 12+\frac{1}{2} \times 12 \times 5\) \(=60+30=90 \mathrm{~m}\)
141188
\(\quad \mathbf{y}=\left(\mathbf{P t}^{2}-\mathbf{Q t}^{3}\right) \mathbf{m}\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
1 \(\frac{27 \mathrm{P}^{3}}{4 \mathrm{Q}^{2}}\)
2 \(\frac{4 Q^{2}}{27 \mathrm{P}^{3}}\)
3 \(\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
4 \(\frac{27 \mathrm{Q}^{2}}{4 \mathrm{P}^{3}}\)
Explanation:
C In vertical plane the vertical displacement of ball is given as - \(\mathrm{y}=\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)\) For maximum height, \(\frac{\mathrm{dy}}{\mathrm{dt}}=0\) \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)=0\) \(\left(2 \mathrm{Pt}-3 \mathrm{Qt}^{2}\right)=0\) \(\mathrm{t}(2 \mathrm{P}-3 \mathrm{Qt})=0\) \(2 \mathrm{P}-3 \mathrm{Qt}=0\) \(\mathrm{t}=\frac{2 \mathrm{P}}{3 \mathrm{Q}}\) Putting the value of \(t\) in equation (i), \(y_{\max } =\frac{4 \mathrm{P}^{3}}{9 \mathrm{Q}^{2}}-\frac{8 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}[3-2]\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
AP EAMCET-04.07.2022
Motion in One Dimensions
141189
\(\quad\) A person moved from \(A\) to \(B\) on a circular path as shown in figure. If the distance travelled by him is \(60 \mathrm{~m}\), then the magnitude of displacement would be: \(\left(\right.\) Given \(\left.\cos 135^{\circ}=-0.7\right)\)
1 \(42 \mathrm{~m}\)
2 \(47 \mathrm{~m}\)
3 \(19 \mathrm{~m}\)
4 \(40 \mathrm{~m}\)
Explanation:
B Distance travelled by Person \(=60 \mathrm{~m}\) (Arc ACB) Let, \(\mathrm{R}=\) Radius Displacement is shortest distance between two point. So, Displacement of person \(=\mathrm{AB}\) (Straight line) Using consines rule in triangle \(\mathrm{BOA}\). \((\mathrm{AB})^{2}=(\mathrm{OA})^{2}+(\mathrm{OB})^{2}-2(\mathrm{OA})(\mathrm{OB}) \cos (\mathrm{BOA})\) \((A B)^{2}=R^{2}+R^{2}-2 R \times R \cos 135^{\circ}\) \((A B)^{2}=2 R^{2}-2 R^{2} \times(-0.7) \quad\left(\because \cos 135^{\circ}=-0.7\right)\) \(\mathrm{AB}=1.84 \mathrm{R}\) Circumference of circle \(=2 \pi R\) \(\mathrm{R}=60 \times \frac{360}{135 \times 2 \pi}=\frac{80}{\pi}\) \(=\frac{80}{22} \times 7\) \(\mathrm{R}=25.45 \mathrm{~m}\) Putting value of \(R\) in \(\mathrm{eq}^{\mathrm{n}}\) (i) we have. \(\mathrm{AB} =1.84 \mathrm{R}=1.84 \times 25.45\) \(\mathrm{AB} =46.92 \mathrm{~m}\) \(\mathrm{AB} \cong 47 \mathrm{~m}\)
JEE Main-25.07.2022
Motion in One Dimensions
141191
The relation between time \(t\) and distance \(x\) for a moving body is given as \(t=\mathbf{m x}^{2}+\mathbf{n x}\), where \(m\) and \(n\) are constants. The retardation of the motion is (when \(v\) stands for velocity)
1 \(2 \mathrm{mv}^{3}\)
2 \(2 \mathrm{mnv}^{3}\)
3 \(2 \mathrm{nv}^{3}\)
4 \(2 n^{2} v^{3}\)
Explanation:
A Given, \(\mathrm{t}=m \mathrm{x}^{2}+\mathrm{nx}\) Differentiating the above equation with respect to time, we have \(\frac{\mathrm{dt}}{\mathrm{dt}}=\mathrm{m} \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=2 \mathrm{mx} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{mxv}+\mathrm{nv}\) \(1=\mathrm{v}(2 \mathrm{mx}+\mathrm{n})\) \(\mathrm{v}=\frac{1}{2 \mathrm{mx}+\mathrm{n}}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{-2 \mathrm{~m}(\mathrm{dx} / \mathrm{dt})}{(2 \mathrm{mx}+\mathrm{n})^{2}}=-2 \mathrm{~m} \times \mathrm{v}^{2} \times \mathrm{v}=-2 \mathrm{mv}^{3}\) \(\therefore\) The retardation of the motion is \(2 \mathrm{mv}^{3}\).
JEE Main-25.07.2021
Motion in One Dimensions
141192
The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between \(t=0 \mathrm{~s}\) and \(t=10\) s will be
1 \(120 \mathrm{~m}\)
2 \(90 \mathrm{~m}\)
3 \(60 \mathrm{~m}\)
4 \(30 \mathrm{~m}\)
Explanation:
B Distance travelled by the particle \(=\) Area under the v-t graph Area of \(\square \mathrm{ABCO}+\) area of \(\triangle \mathrm{BCP}\) \(=5 \times 12+\frac{1}{2} \times 12 \times 5\) \(=60+30=90 \mathrm{~m}\)
141188
\(\quad \mathbf{y}=\left(\mathbf{P t}^{2}-\mathbf{Q t}^{3}\right) \mathbf{m}\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
1 \(\frac{27 \mathrm{P}^{3}}{4 \mathrm{Q}^{2}}\)
2 \(\frac{4 Q^{2}}{27 \mathrm{P}^{3}}\)
3 \(\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
4 \(\frac{27 \mathrm{Q}^{2}}{4 \mathrm{P}^{3}}\)
Explanation:
C In vertical plane the vertical displacement of ball is given as - \(\mathrm{y}=\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)\) For maximum height, \(\frac{\mathrm{dy}}{\mathrm{dt}}=0\) \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{Pt}^{2}-\mathrm{Qt}^{3}\right)=0\) \(\left(2 \mathrm{Pt}-3 \mathrm{Qt}^{2}\right)=0\) \(\mathrm{t}(2 \mathrm{P}-3 \mathrm{Qt})=0\) \(2 \mathrm{P}-3 \mathrm{Qt}=0\) \(\mathrm{t}=\frac{2 \mathrm{P}}{3 \mathrm{Q}}\) Putting the value of \(t\) in equation (i), \(y_{\max } =\frac{4 \mathrm{P}^{3}}{9 \mathrm{Q}^{2}}-\frac{8 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}[3-2]\) \(\mathrm{y}_{\max } =\frac{4 \mathrm{P}^{3}}{27 \mathrm{Q}^{2}}\)
AP EAMCET-04.07.2022
Motion in One Dimensions
141189
\(\quad\) A person moved from \(A\) to \(B\) on a circular path as shown in figure. If the distance travelled by him is \(60 \mathrm{~m}\), then the magnitude of displacement would be: \(\left(\right.\) Given \(\left.\cos 135^{\circ}=-0.7\right)\)
1 \(42 \mathrm{~m}\)
2 \(47 \mathrm{~m}\)
3 \(19 \mathrm{~m}\)
4 \(40 \mathrm{~m}\)
Explanation:
B Distance travelled by Person \(=60 \mathrm{~m}\) (Arc ACB) Let, \(\mathrm{R}=\) Radius Displacement is shortest distance between two point. So, Displacement of person \(=\mathrm{AB}\) (Straight line) Using consines rule in triangle \(\mathrm{BOA}\). \((\mathrm{AB})^{2}=(\mathrm{OA})^{2}+(\mathrm{OB})^{2}-2(\mathrm{OA})(\mathrm{OB}) \cos (\mathrm{BOA})\) \((A B)^{2}=R^{2}+R^{2}-2 R \times R \cos 135^{\circ}\) \((A B)^{2}=2 R^{2}-2 R^{2} \times(-0.7) \quad\left(\because \cos 135^{\circ}=-0.7\right)\) \(\mathrm{AB}=1.84 \mathrm{R}\) Circumference of circle \(=2 \pi R\) \(\mathrm{R}=60 \times \frac{360}{135 \times 2 \pi}=\frac{80}{\pi}\) \(=\frac{80}{22} \times 7\) \(\mathrm{R}=25.45 \mathrm{~m}\) Putting value of \(R\) in \(\mathrm{eq}^{\mathrm{n}}\) (i) we have. \(\mathrm{AB} =1.84 \mathrm{R}=1.84 \times 25.45\) \(\mathrm{AB} =46.92 \mathrm{~m}\) \(\mathrm{AB} \cong 47 \mathrm{~m}\)
JEE Main-25.07.2022
Motion in One Dimensions
141191
The relation between time \(t\) and distance \(x\) for a moving body is given as \(t=\mathbf{m x}^{2}+\mathbf{n x}\), where \(m\) and \(n\) are constants. The retardation of the motion is (when \(v\) stands for velocity)
1 \(2 \mathrm{mv}^{3}\)
2 \(2 \mathrm{mnv}^{3}\)
3 \(2 \mathrm{nv}^{3}\)
4 \(2 n^{2} v^{3}\)
Explanation:
A Given, \(\mathrm{t}=m \mathrm{x}^{2}+\mathrm{nx}\) Differentiating the above equation with respect to time, we have \(\frac{\mathrm{dt}}{\mathrm{dt}}=\mathrm{m} \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=2 \mathrm{mx} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{n} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{mxv}+\mathrm{nv}\) \(1=\mathrm{v}(2 \mathrm{mx}+\mathrm{n})\) \(\mathrm{v}=\frac{1}{2 \mathrm{mx}+\mathrm{n}}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{-2 \mathrm{~m}(\mathrm{dx} / \mathrm{dt})}{(2 \mathrm{mx}+\mathrm{n})^{2}}=-2 \mathrm{~m} \times \mathrm{v}^{2} \times \mathrm{v}=-2 \mathrm{mv}^{3}\) \(\therefore\) The retardation of the motion is \(2 \mathrm{mv}^{3}\).
JEE Main-25.07.2021
Motion in One Dimensions
141192
The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between \(t=0 \mathrm{~s}\) and \(t=10\) s will be
1 \(120 \mathrm{~m}\)
2 \(90 \mathrm{~m}\)
3 \(60 \mathrm{~m}\)
4 \(30 \mathrm{~m}\)
Explanation:
B Distance travelled by the particle \(=\) Area under the v-t graph Area of \(\square \mathrm{ABCO}+\) area of \(\triangle \mathrm{BCP}\) \(=5 \times 12+\frac{1}{2} \times 12 \times 5\) \(=60+30=90 \mathrm{~m}\)
