141361
A particle is moving eastwards with a velocity of \(5 \mathrm{~m} / \mathrm{sec}\). If in \(10 \mathrm{~s}\) the velocity changes by 5 \(\mathrm{m} / \mathrm{s}\) northwards, what is the average acceleration in this time?
A Initial velocity \((\mathrm{u})=5 \mathrm{~m} / \mathrm{s}\) due east Final velocity \((v)=5 \mathrm{~m} / \mathrm{s}\) due north Now, change in velocity \(=\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}}\) \(|\Delta v|=\left(v^{2}+u^{2}\right)^{1 / 2}=\sqrt{5^{2}+5^{2}}\) \(\Rightarrow \Delta \mathrm{v}=5 \sqrt{2} \mathrm{~m} / \mathrm{s}\) Direction of \(\Delta v\) is given by, \(\tan =\frac{\mathrm{v}}{\mathrm{u}}=\frac{5}{-5}=-1\) \(\Rightarrow \theta=-\pi / 4\) Thus, the acceleration is \(\mathrm{a}=\frac{\Delta \mathrm{v}}{\mathrm{t}}=\frac{5 \sqrt{2}}{10}\) \(\mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) and is directed towards North-West direction
MP PMT-2013
Motion in One Dimensions
141362
The relation between the ' \(t\) ' and distance ' \(x\) ' is \(t\) \(=\alpha x^{2}+\beta x\), where \(\alpha \ \beta\) are constants. The retardation is:
141364
A particle is projected with velocity \(V_{0}\) along \(x\) axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. \(a=\alpha x^{2}\), the distance at which the particle stops is
D We know that \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}=-\alpha \mathrm{x}^{2}\) \(\Rightarrow \quad v d v=-\alpha x^{2} d x\) When particle will stop velocity of particle will be zero and it stop at distance (s). \(\therefore \quad \int_{\mathrm{v}_{0}}^{0} \mathrm{vdv}=-\alpha \int_{0}^{\mathrm{s}} \mathrm{x}^{2} \mathrm{dx}\) \(\frac{\mathrm{v}_{0}{ }^{2}}{2}=\frac{\alpha \mathrm{s}^{3}}{3}\) \(\therefore \quad \mathrm{s}=\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \alpha}\right)^{1 / 3}\)
MP PET-2008
Motion in One Dimensions
141365
A particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\) and move towards north for \(8 \mathrm{~s}\) with velocity \(5 \mathrm{~m} / \mathrm{s}\). Then, average velocity of the particle is
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(5 \mathrm{~m} / \mathrm{s}\)
3 \(7 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\). So, displacement of particle in east direction \(=\) velocity \(\mathrm{x}\) time \(\mathrm{s}_{1}=15 \times 2=30 \mathrm{~m}\) When particle move towards north for 8 second with the velocity \(5 \mathrm{~m} / \mathrm{s}\). So displacement of particle in north direction- \(\mathrm{s}_{2}=5 \times 8\) \(\mathrm{~s}_{2}=40 \mathrm{~m}\) Resultant displacement of particle \(=\sqrt{(30)^{2}+(40)^{2}}\) \(\mathrm{s}_{\mathrm{R}}=\sqrt{900+1600}=\sqrt{2500}\) \(\mathrm{~S}_{\mathrm{R}}=50 \mathrm{~m}\) Average velocity \((v)=\frac{S_{R}}{t_{1}+t_{2}}\) \(\mathrm{v}=\frac{50}{2+8}\) \(\mathrm{v}=\frac{50}{10} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\)
141361
A particle is moving eastwards with a velocity of \(5 \mathrm{~m} / \mathrm{sec}\). If in \(10 \mathrm{~s}\) the velocity changes by 5 \(\mathrm{m} / \mathrm{s}\) northwards, what is the average acceleration in this time?
A Initial velocity \((\mathrm{u})=5 \mathrm{~m} / \mathrm{s}\) due east Final velocity \((v)=5 \mathrm{~m} / \mathrm{s}\) due north Now, change in velocity \(=\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}}\) \(|\Delta v|=\left(v^{2}+u^{2}\right)^{1 / 2}=\sqrt{5^{2}+5^{2}}\) \(\Rightarrow \Delta \mathrm{v}=5 \sqrt{2} \mathrm{~m} / \mathrm{s}\) Direction of \(\Delta v\) is given by, \(\tan =\frac{\mathrm{v}}{\mathrm{u}}=\frac{5}{-5}=-1\) \(\Rightarrow \theta=-\pi / 4\) Thus, the acceleration is \(\mathrm{a}=\frac{\Delta \mathrm{v}}{\mathrm{t}}=\frac{5 \sqrt{2}}{10}\) \(\mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) and is directed towards North-West direction
MP PMT-2013
Motion in One Dimensions
141362
The relation between the ' \(t\) ' and distance ' \(x\) ' is \(t\) \(=\alpha x^{2}+\beta x\), where \(\alpha \ \beta\) are constants. The retardation is:
141364
A particle is projected with velocity \(V_{0}\) along \(x\) axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. \(a=\alpha x^{2}\), the distance at which the particle stops is
D We know that \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}=-\alpha \mathrm{x}^{2}\) \(\Rightarrow \quad v d v=-\alpha x^{2} d x\) When particle will stop velocity of particle will be zero and it stop at distance (s). \(\therefore \quad \int_{\mathrm{v}_{0}}^{0} \mathrm{vdv}=-\alpha \int_{0}^{\mathrm{s}} \mathrm{x}^{2} \mathrm{dx}\) \(\frac{\mathrm{v}_{0}{ }^{2}}{2}=\frac{\alpha \mathrm{s}^{3}}{3}\) \(\therefore \quad \mathrm{s}=\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \alpha}\right)^{1 / 3}\)
MP PET-2008
Motion in One Dimensions
141365
A particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\) and move towards north for \(8 \mathrm{~s}\) with velocity \(5 \mathrm{~m} / \mathrm{s}\). Then, average velocity of the particle is
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(5 \mathrm{~m} / \mathrm{s}\)
3 \(7 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\). So, displacement of particle in east direction \(=\) velocity \(\mathrm{x}\) time \(\mathrm{s}_{1}=15 \times 2=30 \mathrm{~m}\) When particle move towards north for 8 second with the velocity \(5 \mathrm{~m} / \mathrm{s}\). So displacement of particle in north direction- \(\mathrm{s}_{2}=5 \times 8\) \(\mathrm{~s}_{2}=40 \mathrm{~m}\) Resultant displacement of particle \(=\sqrt{(30)^{2}+(40)^{2}}\) \(\mathrm{s}_{\mathrm{R}}=\sqrt{900+1600}=\sqrt{2500}\) \(\mathrm{~S}_{\mathrm{R}}=50 \mathrm{~m}\) Average velocity \((v)=\frac{S_{R}}{t_{1}+t_{2}}\) \(\mathrm{v}=\frac{50}{2+8}\) \(\mathrm{v}=\frac{50}{10} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\)
141361
A particle is moving eastwards with a velocity of \(5 \mathrm{~m} / \mathrm{sec}\). If in \(10 \mathrm{~s}\) the velocity changes by 5 \(\mathrm{m} / \mathrm{s}\) northwards, what is the average acceleration in this time?
A Initial velocity \((\mathrm{u})=5 \mathrm{~m} / \mathrm{s}\) due east Final velocity \((v)=5 \mathrm{~m} / \mathrm{s}\) due north Now, change in velocity \(=\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}}\) \(|\Delta v|=\left(v^{2}+u^{2}\right)^{1 / 2}=\sqrt{5^{2}+5^{2}}\) \(\Rightarrow \Delta \mathrm{v}=5 \sqrt{2} \mathrm{~m} / \mathrm{s}\) Direction of \(\Delta v\) is given by, \(\tan =\frac{\mathrm{v}}{\mathrm{u}}=\frac{5}{-5}=-1\) \(\Rightarrow \theta=-\pi / 4\) Thus, the acceleration is \(\mathrm{a}=\frac{\Delta \mathrm{v}}{\mathrm{t}}=\frac{5 \sqrt{2}}{10}\) \(\mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) and is directed towards North-West direction
MP PMT-2013
Motion in One Dimensions
141362
The relation between the ' \(t\) ' and distance ' \(x\) ' is \(t\) \(=\alpha x^{2}+\beta x\), where \(\alpha \ \beta\) are constants. The retardation is:
141364
A particle is projected with velocity \(V_{0}\) along \(x\) axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. \(a=\alpha x^{2}\), the distance at which the particle stops is
D We know that \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}=-\alpha \mathrm{x}^{2}\) \(\Rightarrow \quad v d v=-\alpha x^{2} d x\) When particle will stop velocity of particle will be zero and it stop at distance (s). \(\therefore \quad \int_{\mathrm{v}_{0}}^{0} \mathrm{vdv}=-\alpha \int_{0}^{\mathrm{s}} \mathrm{x}^{2} \mathrm{dx}\) \(\frac{\mathrm{v}_{0}{ }^{2}}{2}=\frac{\alpha \mathrm{s}^{3}}{3}\) \(\therefore \quad \mathrm{s}=\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \alpha}\right)^{1 / 3}\)
MP PET-2008
Motion in One Dimensions
141365
A particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\) and move towards north for \(8 \mathrm{~s}\) with velocity \(5 \mathrm{~m} / \mathrm{s}\). Then, average velocity of the particle is
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(5 \mathrm{~m} / \mathrm{s}\)
3 \(7 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\). So, displacement of particle in east direction \(=\) velocity \(\mathrm{x}\) time \(\mathrm{s}_{1}=15 \times 2=30 \mathrm{~m}\) When particle move towards north for 8 second with the velocity \(5 \mathrm{~m} / \mathrm{s}\). So displacement of particle in north direction- \(\mathrm{s}_{2}=5 \times 8\) \(\mathrm{~s}_{2}=40 \mathrm{~m}\) Resultant displacement of particle \(=\sqrt{(30)^{2}+(40)^{2}}\) \(\mathrm{s}_{\mathrm{R}}=\sqrt{900+1600}=\sqrt{2500}\) \(\mathrm{~S}_{\mathrm{R}}=50 \mathrm{~m}\) Average velocity \((v)=\frac{S_{R}}{t_{1}+t_{2}}\) \(\mathrm{v}=\frac{50}{2+8}\) \(\mathrm{v}=\frac{50}{10} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\)
141361
A particle is moving eastwards with a velocity of \(5 \mathrm{~m} / \mathrm{sec}\). If in \(10 \mathrm{~s}\) the velocity changes by 5 \(\mathrm{m} / \mathrm{s}\) northwards, what is the average acceleration in this time?
A Initial velocity \((\mathrm{u})=5 \mathrm{~m} / \mathrm{s}\) due east Final velocity \((v)=5 \mathrm{~m} / \mathrm{s}\) due north Now, change in velocity \(=\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}}\) \(|\Delta v|=\left(v^{2}+u^{2}\right)^{1 / 2}=\sqrt{5^{2}+5^{2}}\) \(\Rightarrow \Delta \mathrm{v}=5 \sqrt{2} \mathrm{~m} / \mathrm{s}\) Direction of \(\Delta v\) is given by, \(\tan =\frac{\mathrm{v}}{\mathrm{u}}=\frac{5}{-5}=-1\) \(\Rightarrow \theta=-\pi / 4\) Thus, the acceleration is \(\mathrm{a}=\frac{\Delta \mathrm{v}}{\mathrm{t}}=\frac{5 \sqrt{2}}{10}\) \(\mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) and is directed towards North-West direction
MP PMT-2013
Motion in One Dimensions
141362
The relation between the ' \(t\) ' and distance ' \(x\) ' is \(t\) \(=\alpha x^{2}+\beta x\), where \(\alpha \ \beta\) are constants. The retardation is:
141364
A particle is projected with velocity \(V_{0}\) along \(x\) axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. \(a=\alpha x^{2}\), the distance at which the particle stops is
D We know that \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}=-\alpha \mathrm{x}^{2}\) \(\Rightarrow \quad v d v=-\alpha x^{2} d x\) When particle will stop velocity of particle will be zero and it stop at distance (s). \(\therefore \quad \int_{\mathrm{v}_{0}}^{0} \mathrm{vdv}=-\alpha \int_{0}^{\mathrm{s}} \mathrm{x}^{2} \mathrm{dx}\) \(\frac{\mathrm{v}_{0}{ }^{2}}{2}=\frac{\alpha \mathrm{s}^{3}}{3}\) \(\therefore \quad \mathrm{s}=\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \alpha}\right)^{1 / 3}\)
MP PET-2008
Motion in One Dimensions
141365
A particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\) and move towards north for \(8 \mathrm{~s}\) with velocity \(5 \mathrm{~m} / \mathrm{s}\). Then, average velocity of the particle is
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(5 \mathrm{~m} / \mathrm{s}\)
3 \(7 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Average velocity \(=\frac{\text { Total Displacement }}{\text { Total time }}\) Particle moves towards east for \(2 \mathrm{~s}\) with velocity \(15 \mathrm{~m} / \mathrm{s}\). So, displacement of particle in east direction \(=\) velocity \(\mathrm{x}\) time \(\mathrm{s}_{1}=15 \times 2=30 \mathrm{~m}\) When particle move towards north for 8 second with the velocity \(5 \mathrm{~m} / \mathrm{s}\). So displacement of particle in north direction- \(\mathrm{s}_{2}=5 \times 8\) \(\mathrm{~s}_{2}=40 \mathrm{~m}\) Resultant displacement of particle \(=\sqrt{(30)^{2}+(40)^{2}}\) \(\mathrm{s}_{\mathrm{R}}=\sqrt{900+1600}=\sqrt{2500}\) \(\mathrm{~S}_{\mathrm{R}}=50 \mathrm{~m}\) Average velocity \((v)=\frac{S_{R}}{t_{1}+t_{2}}\) \(\mathrm{v}=\frac{50}{2+8}\) \(\mathrm{v}=\frac{50}{10} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\)
