141206
The instantaneous velocity of a particle moving in a straight line is given as \(v=\alpha t+\beta t^{2}\), where \(\alpha\) and \(\beta\) are constants. The distance travelled by the particle between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
1 \(3 \alpha+7 \beta\)
2 \(\frac{3}{2} \alpha+\frac{7}{3} \beta\)
3 \(\frac{\alpha}{2}+\frac{\beta}{3}\)
4 \(\frac{3}{2} \alpha+\frac{7}{2} \beta\)
Explanation:
B The instantaneous velocity of a particle is given as - \(\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(d x=\left(\alpha t+\beta t^{2}\right) d t\) Integrating the above equation from, \(t=1\) to 2 \(\int_{1}^{2} \mathrm{dx}=\int_{1}^{2}[\alpha \mathrm{t}] \mathrm{dt}+\int_{1}^{2} \beta \mathrm{t}^{2} \mathrm{dt}\) \({[\mathrm{x}]_{1}^{2}=\frac{[\alpha \mathrm{t}]_{1}^{2}}{2}+\frac{\left[\beta \mathrm{t}^{3}\right]_{1}^{2}}{3}}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}\left[2^{2}-1^{1}\right]+\frac{\beta}{3}\left[2^{3}-1^{3}\right]}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}[4-1]+\frac{\beta}{3}[8-1]}\) \(\Delta \mathrm{x}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\frac{3 \alpha}{2}+\frac{7 \beta}{3}\)
JEE Main-25.07.2021
Motion in One Dimensions
141207
A time dependent force is acting on a body of which the velocity of body can be expressed as \(\alpha t^{3 / 2}\). The displacement of the body between time \(t=2\) and \(5 \mathrm{sec}\) is about.
1 \(20 \alpha\)
2 \(22 \alpha\)
3 \(6 \alpha\)
4 0
Explanation:
A Velocity of the body is given as - \(v=\alpha t^{3 / 2}\) To find out the displacement between time \((t)=2\) second to 5 second \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}^{3 / 2}\) \(\mathrm{dx}=\alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) Integrating the above equation from \(t=2\) to \(5 s-\) \(\int_{2}^{5} \mathrm{dx}=\int_{2}^{5} \alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) \({\left[\mathrm{x}_{5}-\mathrm{x}_{2}\right]=\alpha\left[\frac{\mathrm{t}^{5 / 2}}{5 / 2}\right]_{2}^{5}}\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}\left[5^{2.5}-2^{2.5}\right]\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}[55.90-5.65]\) \(\Delta \mathrm{x}=\frac{2 \alpha}{5}[50.24]\) \(\Delta \mathrm{x}=20.09 \alpha \approx 20 \alpha\) \(\Delta \mathrm{x}=20 \alpha\)
Assam CEE-2021
Motion in One Dimensions
141208
An object is moving along \(x\)-axis. If its position \(\mathbf{x}(\mathbf{t})\) (in meter) at any time \(t\) (in \(s\) ) is given by the following graph, then the object is at rest about
1 \(t=-1 s\)
2 \(\mathrm{t}=0 \mathrm{~s}\)
3 \(\mathrm{t}=1 \mathrm{~s}\)
4 \(\mathrm{t}=2 \mathrm{~s}\)
Explanation:
C An object is moving along \(x\)-axis. If its position \(x(t)\) in meter at any time \(+(\) in sec) is given by the following graph. Then the object is at rest about \(\mathrm{t}=1 \mathrm{sec}\) Because at \(t=1 \mathrm{sec}\) position of object is not changing.
TS EAMCET 28.09.2020
Motion in One Dimensions
141209
A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of \(72 \mathrm{~km} / \mathrm{hr}\). The angle of inclination of the bike with horizontal plant will be
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(\tan ^{-1}\left(\frac{36}{67}\right)\)
Explanation:
B Radius of circular path \(\mathrm{r}=134 \mathrm{ft}\) \(=134 \times 0.3045 \mathrm{~m}=40.84\) meter Velocity \(\mathrm{V}=72 \mathrm{~km} / \mathrm{hr}\) \(=\frac{72 \times 1000}{3600}=20 \mathrm{~m} / \mathrm{sec}\) The angle of inclination \(\because \tan \theta=\frac{\mathrm{V}^{2}}{\mathrm{rg}}\) \(\tan \theta=\frac{(20)^{2}}{40.84 \times 9.8}=\frac{400}{403.08}=0.9996\) \(\tan \theta \simeq 1\) \(\theta=\tan ^{-1}(1)=45^{\circ}\) Angle of inclination \((\theta)=45^{\circ}\)
141206
The instantaneous velocity of a particle moving in a straight line is given as \(v=\alpha t+\beta t^{2}\), where \(\alpha\) and \(\beta\) are constants. The distance travelled by the particle between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
1 \(3 \alpha+7 \beta\)
2 \(\frac{3}{2} \alpha+\frac{7}{3} \beta\)
3 \(\frac{\alpha}{2}+\frac{\beta}{3}\)
4 \(\frac{3}{2} \alpha+\frac{7}{2} \beta\)
Explanation:
B The instantaneous velocity of a particle is given as - \(\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(d x=\left(\alpha t+\beta t^{2}\right) d t\) Integrating the above equation from, \(t=1\) to 2 \(\int_{1}^{2} \mathrm{dx}=\int_{1}^{2}[\alpha \mathrm{t}] \mathrm{dt}+\int_{1}^{2} \beta \mathrm{t}^{2} \mathrm{dt}\) \({[\mathrm{x}]_{1}^{2}=\frac{[\alpha \mathrm{t}]_{1}^{2}}{2}+\frac{\left[\beta \mathrm{t}^{3}\right]_{1}^{2}}{3}}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}\left[2^{2}-1^{1}\right]+\frac{\beta}{3}\left[2^{3}-1^{3}\right]}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}[4-1]+\frac{\beta}{3}[8-1]}\) \(\Delta \mathrm{x}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\frac{3 \alpha}{2}+\frac{7 \beta}{3}\)
JEE Main-25.07.2021
Motion in One Dimensions
141207
A time dependent force is acting on a body of which the velocity of body can be expressed as \(\alpha t^{3 / 2}\). The displacement of the body between time \(t=2\) and \(5 \mathrm{sec}\) is about.
1 \(20 \alpha\)
2 \(22 \alpha\)
3 \(6 \alpha\)
4 0
Explanation:
A Velocity of the body is given as - \(v=\alpha t^{3 / 2}\) To find out the displacement between time \((t)=2\) second to 5 second \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}^{3 / 2}\) \(\mathrm{dx}=\alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) Integrating the above equation from \(t=2\) to \(5 s-\) \(\int_{2}^{5} \mathrm{dx}=\int_{2}^{5} \alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) \({\left[\mathrm{x}_{5}-\mathrm{x}_{2}\right]=\alpha\left[\frac{\mathrm{t}^{5 / 2}}{5 / 2}\right]_{2}^{5}}\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}\left[5^{2.5}-2^{2.5}\right]\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}[55.90-5.65]\) \(\Delta \mathrm{x}=\frac{2 \alpha}{5}[50.24]\) \(\Delta \mathrm{x}=20.09 \alpha \approx 20 \alpha\) \(\Delta \mathrm{x}=20 \alpha\)
Assam CEE-2021
Motion in One Dimensions
141208
An object is moving along \(x\)-axis. If its position \(\mathbf{x}(\mathbf{t})\) (in meter) at any time \(t\) (in \(s\) ) is given by the following graph, then the object is at rest about
1 \(t=-1 s\)
2 \(\mathrm{t}=0 \mathrm{~s}\)
3 \(\mathrm{t}=1 \mathrm{~s}\)
4 \(\mathrm{t}=2 \mathrm{~s}\)
Explanation:
C An object is moving along \(x\)-axis. If its position \(x(t)\) in meter at any time \(+(\) in sec) is given by the following graph. Then the object is at rest about \(\mathrm{t}=1 \mathrm{sec}\) Because at \(t=1 \mathrm{sec}\) position of object is not changing.
TS EAMCET 28.09.2020
Motion in One Dimensions
141209
A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of \(72 \mathrm{~km} / \mathrm{hr}\). The angle of inclination of the bike with horizontal plant will be
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(\tan ^{-1}\left(\frac{36}{67}\right)\)
Explanation:
B Radius of circular path \(\mathrm{r}=134 \mathrm{ft}\) \(=134 \times 0.3045 \mathrm{~m}=40.84\) meter Velocity \(\mathrm{V}=72 \mathrm{~km} / \mathrm{hr}\) \(=\frac{72 \times 1000}{3600}=20 \mathrm{~m} / \mathrm{sec}\) The angle of inclination \(\because \tan \theta=\frac{\mathrm{V}^{2}}{\mathrm{rg}}\) \(\tan \theta=\frac{(20)^{2}}{40.84 \times 9.8}=\frac{400}{403.08}=0.9996\) \(\tan \theta \simeq 1\) \(\theta=\tan ^{-1}(1)=45^{\circ}\) Angle of inclination \((\theta)=45^{\circ}\)
141206
The instantaneous velocity of a particle moving in a straight line is given as \(v=\alpha t+\beta t^{2}\), where \(\alpha\) and \(\beta\) are constants. The distance travelled by the particle between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
1 \(3 \alpha+7 \beta\)
2 \(\frac{3}{2} \alpha+\frac{7}{3} \beta\)
3 \(\frac{\alpha}{2}+\frac{\beta}{3}\)
4 \(\frac{3}{2} \alpha+\frac{7}{2} \beta\)
Explanation:
B The instantaneous velocity of a particle is given as - \(\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(d x=\left(\alpha t+\beta t^{2}\right) d t\) Integrating the above equation from, \(t=1\) to 2 \(\int_{1}^{2} \mathrm{dx}=\int_{1}^{2}[\alpha \mathrm{t}] \mathrm{dt}+\int_{1}^{2} \beta \mathrm{t}^{2} \mathrm{dt}\) \({[\mathrm{x}]_{1}^{2}=\frac{[\alpha \mathrm{t}]_{1}^{2}}{2}+\frac{\left[\beta \mathrm{t}^{3}\right]_{1}^{2}}{3}}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}\left[2^{2}-1^{1}\right]+\frac{\beta}{3}\left[2^{3}-1^{3}\right]}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}[4-1]+\frac{\beta}{3}[8-1]}\) \(\Delta \mathrm{x}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\frac{3 \alpha}{2}+\frac{7 \beta}{3}\)
JEE Main-25.07.2021
Motion in One Dimensions
141207
A time dependent force is acting on a body of which the velocity of body can be expressed as \(\alpha t^{3 / 2}\). The displacement of the body between time \(t=2\) and \(5 \mathrm{sec}\) is about.
1 \(20 \alpha\)
2 \(22 \alpha\)
3 \(6 \alpha\)
4 0
Explanation:
A Velocity of the body is given as - \(v=\alpha t^{3 / 2}\) To find out the displacement between time \((t)=2\) second to 5 second \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}^{3 / 2}\) \(\mathrm{dx}=\alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) Integrating the above equation from \(t=2\) to \(5 s-\) \(\int_{2}^{5} \mathrm{dx}=\int_{2}^{5} \alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) \({\left[\mathrm{x}_{5}-\mathrm{x}_{2}\right]=\alpha\left[\frac{\mathrm{t}^{5 / 2}}{5 / 2}\right]_{2}^{5}}\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}\left[5^{2.5}-2^{2.5}\right]\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}[55.90-5.65]\) \(\Delta \mathrm{x}=\frac{2 \alpha}{5}[50.24]\) \(\Delta \mathrm{x}=20.09 \alpha \approx 20 \alpha\) \(\Delta \mathrm{x}=20 \alpha\)
Assam CEE-2021
Motion in One Dimensions
141208
An object is moving along \(x\)-axis. If its position \(\mathbf{x}(\mathbf{t})\) (in meter) at any time \(t\) (in \(s\) ) is given by the following graph, then the object is at rest about
1 \(t=-1 s\)
2 \(\mathrm{t}=0 \mathrm{~s}\)
3 \(\mathrm{t}=1 \mathrm{~s}\)
4 \(\mathrm{t}=2 \mathrm{~s}\)
Explanation:
C An object is moving along \(x\)-axis. If its position \(x(t)\) in meter at any time \(+(\) in sec) is given by the following graph. Then the object is at rest about \(\mathrm{t}=1 \mathrm{sec}\) Because at \(t=1 \mathrm{sec}\) position of object is not changing.
TS EAMCET 28.09.2020
Motion in One Dimensions
141209
A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of \(72 \mathrm{~km} / \mathrm{hr}\). The angle of inclination of the bike with horizontal plant will be
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(\tan ^{-1}\left(\frac{36}{67}\right)\)
Explanation:
B Radius of circular path \(\mathrm{r}=134 \mathrm{ft}\) \(=134 \times 0.3045 \mathrm{~m}=40.84\) meter Velocity \(\mathrm{V}=72 \mathrm{~km} / \mathrm{hr}\) \(=\frac{72 \times 1000}{3600}=20 \mathrm{~m} / \mathrm{sec}\) The angle of inclination \(\because \tan \theta=\frac{\mathrm{V}^{2}}{\mathrm{rg}}\) \(\tan \theta=\frac{(20)^{2}}{40.84 \times 9.8}=\frac{400}{403.08}=0.9996\) \(\tan \theta \simeq 1\) \(\theta=\tan ^{-1}(1)=45^{\circ}\) Angle of inclination \((\theta)=45^{\circ}\)
141206
The instantaneous velocity of a particle moving in a straight line is given as \(v=\alpha t+\beta t^{2}\), where \(\alpha\) and \(\beta\) are constants. The distance travelled by the particle between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
1 \(3 \alpha+7 \beta\)
2 \(\frac{3}{2} \alpha+\frac{7}{3} \beta\)
3 \(\frac{\alpha}{2}+\frac{\beta}{3}\)
4 \(\frac{3}{2} \alpha+\frac{7}{2} \beta\)
Explanation:
B The instantaneous velocity of a particle is given as - \(\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(\because \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}\) \(d x=\left(\alpha t+\beta t^{2}\right) d t\) Integrating the above equation from, \(t=1\) to 2 \(\int_{1}^{2} \mathrm{dx}=\int_{1}^{2}[\alpha \mathrm{t}] \mathrm{dt}+\int_{1}^{2} \beta \mathrm{t}^{2} \mathrm{dt}\) \({[\mathrm{x}]_{1}^{2}=\frac{[\alpha \mathrm{t}]_{1}^{2}}{2}+\frac{\left[\beta \mathrm{t}^{3}\right]_{1}^{2}}{3}}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}\left[2^{2}-1^{1}\right]+\frac{\beta}{3}\left[2^{3}-1^{3}\right]}\) \({[\mathrm{x}]_{1}^{2}=\frac{\alpha}{2}[4-1]+\frac{\beta}{3}[8-1]}\) \(\Delta \mathrm{x}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\frac{3 \alpha}{2}+\frac{7 \beta}{3}\)
JEE Main-25.07.2021
Motion in One Dimensions
141207
A time dependent force is acting on a body of which the velocity of body can be expressed as \(\alpha t^{3 / 2}\). The displacement of the body between time \(t=2\) and \(5 \mathrm{sec}\) is about.
1 \(20 \alpha\)
2 \(22 \alpha\)
3 \(6 \alpha\)
4 0
Explanation:
A Velocity of the body is given as - \(v=\alpha t^{3 / 2}\) To find out the displacement between time \((t)=2\) second to 5 second \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathrm{t}^{3 / 2}\) \(\mathrm{dx}=\alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) Integrating the above equation from \(t=2\) to \(5 s-\) \(\int_{2}^{5} \mathrm{dx}=\int_{2}^{5} \alpha \mathrm{t}^{3 / 2} \mathrm{dt}\) \({\left[\mathrm{x}_{5}-\mathrm{x}_{2}\right]=\alpha\left[\frac{\mathrm{t}^{5 / 2}}{5 / 2}\right]_{2}^{5}}\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}\left[5^{2.5}-2^{2.5}\right]\) \(\Delta \mathrm{x}=\alpha \times \frac{2}{5}[55.90-5.65]\) \(\Delta \mathrm{x}=\frac{2 \alpha}{5}[50.24]\) \(\Delta \mathrm{x}=20.09 \alpha \approx 20 \alpha\) \(\Delta \mathrm{x}=20 \alpha\)
Assam CEE-2021
Motion in One Dimensions
141208
An object is moving along \(x\)-axis. If its position \(\mathbf{x}(\mathbf{t})\) (in meter) at any time \(t\) (in \(s\) ) is given by the following graph, then the object is at rest about
1 \(t=-1 s\)
2 \(\mathrm{t}=0 \mathrm{~s}\)
3 \(\mathrm{t}=1 \mathrm{~s}\)
4 \(\mathrm{t}=2 \mathrm{~s}\)
Explanation:
C An object is moving along \(x\)-axis. If its position \(x(t)\) in meter at any time \(+(\) in sec) is given by the following graph. Then the object is at rest about \(\mathrm{t}=1 \mathrm{sec}\) Because at \(t=1 \mathrm{sec}\) position of object is not changing.
TS EAMCET 28.09.2020
Motion in One Dimensions
141209
A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of \(72 \mathrm{~km} / \mathrm{hr}\). The angle of inclination of the bike with horizontal plant will be
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(\tan ^{-1}\left(\frac{36}{67}\right)\)
Explanation:
B Radius of circular path \(\mathrm{r}=134 \mathrm{ft}\) \(=134 \times 0.3045 \mathrm{~m}=40.84\) meter Velocity \(\mathrm{V}=72 \mathrm{~km} / \mathrm{hr}\) \(=\frac{72 \times 1000}{3600}=20 \mathrm{~m} / \mathrm{sec}\) The angle of inclination \(\because \tan \theta=\frac{\mathrm{V}^{2}}{\mathrm{rg}}\) \(\tan \theta=\frac{(20)^{2}}{40.84 \times 9.8}=\frac{400}{403.08}=0.9996\) \(\tan \theta \simeq 1\) \(\theta=\tan ^{-1}(1)=45^{\circ}\) Angle of inclination \((\theta)=45^{\circ}\)
