141332
A car moves from \(X\) to \(Y\) with a uniform speed \(v_{u}\) and returns to \(Y\) with a uniform speed \(v_{d}\). The average speed for this round trip is
1 \(\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}\)
2 \(\sqrt{v_{u} v_{d}}\)
3 \(\frac{v_{d} v_{u}}{v_{d}+v_{u}}\)
4 \(\frac{v_{u}+v_{d}}{2}\)
Explanation:
A \(\mathrm{X} \mathrm{F}^{\mathrm{V}_{\mathrm{u}}} \mathrm{d}\) \(\therefore \mathrm{t}=\frac{\text { distance }}{\text { time }}\) Time taken by car from \(\mathrm{X}\) to \(\mathrm{Y}\) \(\mathrm{t}_{1}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}\) Time taken by car from \(\mathrm{Y}\) to \(\mathrm{X}\) \(\mathrm{t}_{2}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{\mathrm{d}+\mathrm{d}}{\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}+\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}}\) \(=\frac{2 \mathrm{~d}}{\mathrm{~d}\left(\frac{1}{\mathrm{v}_{\mathrm{u}}}+\frac{1}{\mathrm{v}_{\mathrm{d}}}\right)}\) Average speed, \(V_{a v g}=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}\)
UP CPMT-2008
Motion in One Dimensions
141333
A boat is sent across a river with a velocity of 8 \(\mathbf{k m} / \mathrm{h}\). If the resultant velocity of boat is 10 \(\mathrm{km} / \mathrm{h}\), then velocity of the river is
1 \(10 \mathrm{~km} / \mathrm{h}\)
2 \(8 \mathrm{~km} / \mathrm{h}\)
3 \(6 \mathrm{~km} / \mathrm{h}\)
4 \(4 \mathrm{~km} / \mathrm{h}\)
Explanation:
C Velocity of boat \(\mathrm{v}_{\mathrm{b}}=8 \mathrm{~km} / \mathrm{hr}\) Resultant Velocity \(\left(\mathrm{v}_{\mathrm{r}}\right)=10 \mathrm{~km} / \mathrm{hr}\) \(\mathrm{v}_{\mathrm{r}}^{2}=\mathrm{u}^{2}+\mathrm{v}_{\mathrm{b}}^{2}\) Velocity of river \((u)=\sqrt{v_{r}^{2}-v_{b}^{2}}\) \(=\sqrt{(10)^{2}-(8)^{2}}=6 \mathrm{~km} / \mathrm{hr}\)
UP CPMT-2010
Motion in One Dimensions
141334
A point initially at rest moves along \(x\)-axis. Its acceleration varies with time as \(a=(6 t+5)\) \(\mathrm{m} / \mathrm{s}^{2}\). If it starts from origin, the distance covered in \(2 \mathrm{~s}\) is
1 \(20 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(16 \mathrm{~m}\)
4 \(25 \mathrm{~m}\)
Explanation:
B Given, the acceleration of the particle varies with time as; \(a=(6 t+5) \mathrm{m} / \mathrm{s}^{2}\) Integrating the equation (i), we get- \(\int(6 t+5) d t=v\) \(\frac{6 t^{2}}{2}+5 t=v\) \(3 t^{2}+5 t=v\) Now, given that the velocity of the particle varies with time as \(v=\left(3 t^{2}+5 t\right) \mathrm{ms}^{-1}\). Integrating the equation, \(\int\left(3 \mathrm{t}^{2}+5 \mathrm{t}\right) \mathrm{dt}=\mathrm{x}\) \(\frac{3 \mathrm{t}^{3}}{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x} \Rightarrow \mathrm{t}^{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x}\) At, \(\quad \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{x}=2^{3}+\frac{5 \times 2^{2}}{2}=18 \mathrm{~m}\)
UP CPMT-2006
Motion in One Dimensions
141335
If relation between distance and time is \(s=a+b t+c t^{2}\), find initial velocity and acceleration.
1 \(b+2 c t, 2 c\)
2 b, 2c
3 \(2 \mathrm{c}, \mathrm{b}\)
4 \(b+2 c, 2 c\)
Explanation:
B Given, if relation between distance and time is \(s=a+b t+c t^{2}\) Differentiation both sides w.r.t time, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0+\mathrm{b}+2 \mathrm{ct}=\mathrm{b}+2 \mathrm{ct}\) Initial velocity \(=\left|\frac{\mathrm{ds}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=\mathrm{b}+2 \mathrm{c} \times 0\) \(=\mathrm{b}+0=\mathrm{b}\) \(\because \quad\) Acceleration \((a)=\frac{d v}{d t}=2 c\) Initial acceleration \(\left|\frac{\mathrm{dv}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=2 \mathrm{c}\) Hence, initial velocity \(=\mathrm{b}\) and initial acceleration \(=2 \mathrm{c}\)
UP CPMT-2003
Motion in One Dimensions
141336
A ball is thrown upwards, it takes \(4 \mathrm{~s}\) to reach back to the ground. Find its initial velocity
1 \(30 \mathrm{~ms}^{-1}\)
2 \(10 \mathrm{~ms}^{-1}\)
3 \(40 \mathrm{~ms}^{-1}\)
4 \(20 \mathrm{~ms}^{-1}\)
Explanation:
D Given that, Time \((\mathrm{t})=4 \mathrm{sec}(\) for whole trip \()\) \(\therefore\) Time to reach maximum height is half of total time \(\mathrm{t}_{\max }=\frac{4}{2}=2 \mathrm{sec}\) Also speed at maximum height is zero, \(\therefore \quad\) Final velocity \((\mathrm{v})=0\) We know that, \(\mathrm{v}=\mathrm{u}-\mathrm{g}_{\max }\) \(\mathrm{u}=0+10 \times 2=20 \mathrm{~m} / \mathrm{s}\)
141332
A car moves from \(X\) to \(Y\) with a uniform speed \(v_{u}\) and returns to \(Y\) with a uniform speed \(v_{d}\). The average speed for this round trip is
1 \(\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}\)
2 \(\sqrt{v_{u} v_{d}}\)
3 \(\frac{v_{d} v_{u}}{v_{d}+v_{u}}\)
4 \(\frac{v_{u}+v_{d}}{2}\)
Explanation:
A \(\mathrm{X} \mathrm{F}^{\mathrm{V}_{\mathrm{u}}} \mathrm{d}\) \(\therefore \mathrm{t}=\frac{\text { distance }}{\text { time }}\) Time taken by car from \(\mathrm{X}\) to \(\mathrm{Y}\) \(\mathrm{t}_{1}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}\) Time taken by car from \(\mathrm{Y}\) to \(\mathrm{X}\) \(\mathrm{t}_{2}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{\mathrm{d}+\mathrm{d}}{\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}+\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}}\) \(=\frac{2 \mathrm{~d}}{\mathrm{~d}\left(\frac{1}{\mathrm{v}_{\mathrm{u}}}+\frac{1}{\mathrm{v}_{\mathrm{d}}}\right)}\) Average speed, \(V_{a v g}=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}\)
UP CPMT-2008
Motion in One Dimensions
141333
A boat is sent across a river with a velocity of 8 \(\mathbf{k m} / \mathrm{h}\). If the resultant velocity of boat is 10 \(\mathrm{km} / \mathrm{h}\), then velocity of the river is
1 \(10 \mathrm{~km} / \mathrm{h}\)
2 \(8 \mathrm{~km} / \mathrm{h}\)
3 \(6 \mathrm{~km} / \mathrm{h}\)
4 \(4 \mathrm{~km} / \mathrm{h}\)
Explanation:
C Velocity of boat \(\mathrm{v}_{\mathrm{b}}=8 \mathrm{~km} / \mathrm{hr}\) Resultant Velocity \(\left(\mathrm{v}_{\mathrm{r}}\right)=10 \mathrm{~km} / \mathrm{hr}\) \(\mathrm{v}_{\mathrm{r}}^{2}=\mathrm{u}^{2}+\mathrm{v}_{\mathrm{b}}^{2}\) Velocity of river \((u)=\sqrt{v_{r}^{2}-v_{b}^{2}}\) \(=\sqrt{(10)^{2}-(8)^{2}}=6 \mathrm{~km} / \mathrm{hr}\)
UP CPMT-2010
Motion in One Dimensions
141334
A point initially at rest moves along \(x\)-axis. Its acceleration varies with time as \(a=(6 t+5)\) \(\mathrm{m} / \mathrm{s}^{2}\). If it starts from origin, the distance covered in \(2 \mathrm{~s}\) is
1 \(20 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(16 \mathrm{~m}\)
4 \(25 \mathrm{~m}\)
Explanation:
B Given, the acceleration of the particle varies with time as; \(a=(6 t+5) \mathrm{m} / \mathrm{s}^{2}\) Integrating the equation (i), we get- \(\int(6 t+5) d t=v\) \(\frac{6 t^{2}}{2}+5 t=v\) \(3 t^{2}+5 t=v\) Now, given that the velocity of the particle varies with time as \(v=\left(3 t^{2}+5 t\right) \mathrm{ms}^{-1}\). Integrating the equation, \(\int\left(3 \mathrm{t}^{2}+5 \mathrm{t}\right) \mathrm{dt}=\mathrm{x}\) \(\frac{3 \mathrm{t}^{3}}{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x} \Rightarrow \mathrm{t}^{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x}\) At, \(\quad \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{x}=2^{3}+\frac{5 \times 2^{2}}{2}=18 \mathrm{~m}\)
UP CPMT-2006
Motion in One Dimensions
141335
If relation between distance and time is \(s=a+b t+c t^{2}\), find initial velocity and acceleration.
1 \(b+2 c t, 2 c\)
2 b, 2c
3 \(2 \mathrm{c}, \mathrm{b}\)
4 \(b+2 c, 2 c\)
Explanation:
B Given, if relation between distance and time is \(s=a+b t+c t^{2}\) Differentiation both sides w.r.t time, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0+\mathrm{b}+2 \mathrm{ct}=\mathrm{b}+2 \mathrm{ct}\) Initial velocity \(=\left|\frac{\mathrm{ds}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=\mathrm{b}+2 \mathrm{c} \times 0\) \(=\mathrm{b}+0=\mathrm{b}\) \(\because \quad\) Acceleration \((a)=\frac{d v}{d t}=2 c\) Initial acceleration \(\left|\frac{\mathrm{dv}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=2 \mathrm{c}\) Hence, initial velocity \(=\mathrm{b}\) and initial acceleration \(=2 \mathrm{c}\)
UP CPMT-2003
Motion in One Dimensions
141336
A ball is thrown upwards, it takes \(4 \mathrm{~s}\) to reach back to the ground. Find its initial velocity
1 \(30 \mathrm{~ms}^{-1}\)
2 \(10 \mathrm{~ms}^{-1}\)
3 \(40 \mathrm{~ms}^{-1}\)
4 \(20 \mathrm{~ms}^{-1}\)
Explanation:
D Given that, Time \((\mathrm{t})=4 \mathrm{sec}(\) for whole trip \()\) \(\therefore\) Time to reach maximum height is half of total time \(\mathrm{t}_{\max }=\frac{4}{2}=2 \mathrm{sec}\) Also speed at maximum height is zero, \(\therefore \quad\) Final velocity \((\mathrm{v})=0\) We know that, \(\mathrm{v}=\mathrm{u}-\mathrm{g}_{\max }\) \(\mathrm{u}=0+10 \times 2=20 \mathrm{~m} / \mathrm{s}\)
141332
A car moves from \(X\) to \(Y\) with a uniform speed \(v_{u}\) and returns to \(Y\) with a uniform speed \(v_{d}\). The average speed for this round trip is
1 \(\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}\)
2 \(\sqrt{v_{u} v_{d}}\)
3 \(\frac{v_{d} v_{u}}{v_{d}+v_{u}}\)
4 \(\frac{v_{u}+v_{d}}{2}\)
Explanation:
A \(\mathrm{X} \mathrm{F}^{\mathrm{V}_{\mathrm{u}}} \mathrm{d}\) \(\therefore \mathrm{t}=\frac{\text { distance }}{\text { time }}\) Time taken by car from \(\mathrm{X}\) to \(\mathrm{Y}\) \(\mathrm{t}_{1}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}\) Time taken by car from \(\mathrm{Y}\) to \(\mathrm{X}\) \(\mathrm{t}_{2}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{\mathrm{d}+\mathrm{d}}{\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}+\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}}\) \(=\frac{2 \mathrm{~d}}{\mathrm{~d}\left(\frac{1}{\mathrm{v}_{\mathrm{u}}}+\frac{1}{\mathrm{v}_{\mathrm{d}}}\right)}\) Average speed, \(V_{a v g}=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}\)
UP CPMT-2008
Motion in One Dimensions
141333
A boat is sent across a river with a velocity of 8 \(\mathbf{k m} / \mathrm{h}\). If the resultant velocity of boat is 10 \(\mathrm{km} / \mathrm{h}\), then velocity of the river is
1 \(10 \mathrm{~km} / \mathrm{h}\)
2 \(8 \mathrm{~km} / \mathrm{h}\)
3 \(6 \mathrm{~km} / \mathrm{h}\)
4 \(4 \mathrm{~km} / \mathrm{h}\)
Explanation:
C Velocity of boat \(\mathrm{v}_{\mathrm{b}}=8 \mathrm{~km} / \mathrm{hr}\) Resultant Velocity \(\left(\mathrm{v}_{\mathrm{r}}\right)=10 \mathrm{~km} / \mathrm{hr}\) \(\mathrm{v}_{\mathrm{r}}^{2}=\mathrm{u}^{2}+\mathrm{v}_{\mathrm{b}}^{2}\) Velocity of river \((u)=\sqrt{v_{r}^{2}-v_{b}^{2}}\) \(=\sqrt{(10)^{2}-(8)^{2}}=6 \mathrm{~km} / \mathrm{hr}\)
UP CPMT-2010
Motion in One Dimensions
141334
A point initially at rest moves along \(x\)-axis. Its acceleration varies with time as \(a=(6 t+5)\) \(\mathrm{m} / \mathrm{s}^{2}\). If it starts from origin, the distance covered in \(2 \mathrm{~s}\) is
1 \(20 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(16 \mathrm{~m}\)
4 \(25 \mathrm{~m}\)
Explanation:
B Given, the acceleration of the particle varies with time as; \(a=(6 t+5) \mathrm{m} / \mathrm{s}^{2}\) Integrating the equation (i), we get- \(\int(6 t+5) d t=v\) \(\frac{6 t^{2}}{2}+5 t=v\) \(3 t^{2}+5 t=v\) Now, given that the velocity of the particle varies with time as \(v=\left(3 t^{2}+5 t\right) \mathrm{ms}^{-1}\). Integrating the equation, \(\int\left(3 \mathrm{t}^{2}+5 \mathrm{t}\right) \mathrm{dt}=\mathrm{x}\) \(\frac{3 \mathrm{t}^{3}}{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x} \Rightarrow \mathrm{t}^{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x}\) At, \(\quad \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{x}=2^{3}+\frac{5 \times 2^{2}}{2}=18 \mathrm{~m}\)
UP CPMT-2006
Motion in One Dimensions
141335
If relation between distance and time is \(s=a+b t+c t^{2}\), find initial velocity and acceleration.
1 \(b+2 c t, 2 c\)
2 b, 2c
3 \(2 \mathrm{c}, \mathrm{b}\)
4 \(b+2 c, 2 c\)
Explanation:
B Given, if relation between distance and time is \(s=a+b t+c t^{2}\) Differentiation both sides w.r.t time, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0+\mathrm{b}+2 \mathrm{ct}=\mathrm{b}+2 \mathrm{ct}\) Initial velocity \(=\left|\frac{\mathrm{ds}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=\mathrm{b}+2 \mathrm{c} \times 0\) \(=\mathrm{b}+0=\mathrm{b}\) \(\because \quad\) Acceleration \((a)=\frac{d v}{d t}=2 c\) Initial acceleration \(\left|\frac{\mathrm{dv}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=2 \mathrm{c}\) Hence, initial velocity \(=\mathrm{b}\) and initial acceleration \(=2 \mathrm{c}\)
UP CPMT-2003
Motion in One Dimensions
141336
A ball is thrown upwards, it takes \(4 \mathrm{~s}\) to reach back to the ground. Find its initial velocity
1 \(30 \mathrm{~ms}^{-1}\)
2 \(10 \mathrm{~ms}^{-1}\)
3 \(40 \mathrm{~ms}^{-1}\)
4 \(20 \mathrm{~ms}^{-1}\)
Explanation:
D Given that, Time \((\mathrm{t})=4 \mathrm{sec}(\) for whole trip \()\) \(\therefore\) Time to reach maximum height is half of total time \(\mathrm{t}_{\max }=\frac{4}{2}=2 \mathrm{sec}\) Also speed at maximum height is zero, \(\therefore \quad\) Final velocity \((\mathrm{v})=0\) We know that, \(\mathrm{v}=\mathrm{u}-\mathrm{g}_{\max }\) \(\mathrm{u}=0+10 \times 2=20 \mathrm{~m} / \mathrm{s}\)
141332
A car moves from \(X\) to \(Y\) with a uniform speed \(v_{u}\) and returns to \(Y\) with a uniform speed \(v_{d}\). The average speed for this round trip is
1 \(\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}\)
2 \(\sqrt{v_{u} v_{d}}\)
3 \(\frac{v_{d} v_{u}}{v_{d}+v_{u}}\)
4 \(\frac{v_{u}+v_{d}}{2}\)
Explanation:
A \(\mathrm{X} \mathrm{F}^{\mathrm{V}_{\mathrm{u}}} \mathrm{d}\) \(\therefore \mathrm{t}=\frac{\text { distance }}{\text { time }}\) Time taken by car from \(\mathrm{X}\) to \(\mathrm{Y}\) \(\mathrm{t}_{1}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}\) Time taken by car from \(\mathrm{Y}\) to \(\mathrm{X}\) \(\mathrm{t}_{2}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{\mathrm{d}+\mathrm{d}}{\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}+\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}}\) \(=\frac{2 \mathrm{~d}}{\mathrm{~d}\left(\frac{1}{\mathrm{v}_{\mathrm{u}}}+\frac{1}{\mathrm{v}_{\mathrm{d}}}\right)}\) Average speed, \(V_{a v g}=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}\)
UP CPMT-2008
Motion in One Dimensions
141333
A boat is sent across a river with a velocity of 8 \(\mathbf{k m} / \mathrm{h}\). If the resultant velocity of boat is 10 \(\mathrm{km} / \mathrm{h}\), then velocity of the river is
1 \(10 \mathrm{~km} / \mathrm{h}\)
2 \(8 \mathrm{~km} / \mathrm{h}\)
3 \(6 \mathrm{~km} / \mathrm{h}\)
4 \(4 \mathrm{~km} / \mathrm{h}\)
Explanation:
C Velocity of boat \(\mathrm{v}_{\mathrm{b}}=8 \mathrm{~km} / \mathrm{hr}\) Resultant Velocity \(\left(\mathrm{v}_{\mathrm{r}}\right)=10 \mathrm{~km} / \mathrm{hr}\) \(\mathrm{v}_{\mathrm{r}}^{2}=\mathrm{u}^{2}+\mathrm{v}_{\mathrm{b}}^{2}\) Velocity of river \((u)=\sqrt{v_{r}^{2}-v_{b}^{2}}\) \(=\sqrt{(10)^{2}-(8)^{2}}=6 \mathrm{~km} / \mathrm{hr}\)
UP CPMT-2010
Motion in One Dimensions
141334
A point initially at rest moves along \(x\)-axis. Its acceleration varies with time as \(a=(6 t+5)\) \(\mathrm{m} / \mathrm{s}^{2}\). If it starts from origin, the distance covered in \(2 \mathrm{~s}\) is
1 \(20 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(16 \mathrm{~m}\)
4 \(25 \mathrm{~m}\)
Explanation:
B Given, the acceleration of the particle varies with time as; \(a=(6 t+5) \mathrm{m} / \mathrm{s}^{2}\) Integrating the equation (i), we get- \(\int(6 t+5) d t=v\) \(\frac{6 t^{2}}{2}+5 t=v\) \(3 t^{2}+5 t=v\) Now, given that the velocity of the particle varies with time as \(v=\left(3 t^{2}+5 t\right) \mathrm{ms}^{-1}\). Integrating the equation, \(\int\left(3 \mathrm{t}^{2}+5 \mathrm{t}\right) \mathrm{dt}=\mathrm{x}\) \(\frac{3 \mathrm{t}^{3}}{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x} \Rightarrow \mathrm{t}^{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x}\) At, \(\quad \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{x}=2^{3}+\frac{5 \times 2^{2}}{2}=18 \mathrm{~m}\)
UP CPMT-2006
Motion in One Dimensions
141335
If relation between distance and time is \(s=a+b t+c t^{2}\), find initial velocity and acceleration.
1 \(b+2 c t, 2 c\)
2 b, 2c
3 \(2 \mathrm{c}, \mathrm{b}\)
4 \(b+2 c, 2 c\)
Explanation:
B Given, if relation between distance and time is \(s=a+b t+c t^{2}\) Differentiation both sides w.r.t time, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0+\mathrm{b}+2 \mathrm{ct}=\mathrm{b}+2 \mathrm{ct}\) Initial velocity \(=\left|\frac{\mathrm{ds}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=\mathrm{b}+2 \mathrm{c} \times 0\) \(=\mathrm{b}+0=\mathrm{b}\) \(\because \quad\) Acceleration \((a)=\frac{d v}{d t}=2 c\) Initial acceleration \(\left|\frac{\mathrm{dv}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=2 \mathrm{c}\) Hence, initial velocity \(=\mathrm{b}\) and initial acceleration \(=2 \mathrm{c}\)
UP CPMT-2003
Motion in One Dimensions
141336
A ball is thrown upwards, it takes \(4 \mathrm{~s}\) to reach back to the ground. Find its initial velocity
1 \(30 \mathrm{~ms}^{-1}\)
2 \(10 \mathrm{~ms}^{-1}\)
3 \(40 \mathrm{~ms}^{-1}\)
4 \(20 \mathrm{~ms}^{-1}\)
Explanation:
D Given that, Time \((\mathrm{t})=4 \mathrm{sec}(\) for whole trip \()\) \(\therefore\) Time to reach maximum height is half of total time \(\mathrm{t}_{\max }=\frac{4}{2}=2 \mathrm{sec}\) Also speed at maximum height is zero, \(\therefore \quad\) Final velocity \((\mathrm{v})=0\) We know that, \(\mathrm{v}=\mathrm{u}-\mathrm{g}_{\max }\) \(\mathrm{u}=0+10 \times 2=20 \mathrm{~m} / \mathrm{s}\)
141332
A car moves from \(X\) to \(Y\) with a uniform speed \(v_{u}\) and returns to \(Y\) with a uniform speed \(v_{d}\). The average speed for this round trip is
1 \(\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}\)
2 \(\sqrt{v_{u} v_{d}}\)
3 \(\frac{v_{d} v_{u}}{v_{d}+v_{u}}\)
4 \(\frac{v_{u}+v_{d}}{2}\)
Explanation:
A \(\mathrm{X} \mathrm{F}^{\mathrm{V}_{\mathrm{u}}} \mathrm{d}\) \(\therefore \mathrm{t}=\frac{\text { distance }}{\text { time }}\) Time taken by car from \(\mathrm{X}\) to \(\mathrm{Y}\) \(\mathrm{t}_{1}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}\) Time taken by car from \(\mathrm{Y}\) to \(\mathrm{X}\) \(\mathrm{t}_{2}=\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{\mathrm{d}+\mathrm{d}}{\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{u}}}+\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{d}}}}\) \(=\frac{2 \mathrm{~d}}{\mathrm{~d}\left(\frac{1}{\mathrm{v}_{\mathrm{u}}}+\frac{1}{\mathrm{v}_{\mathrm{d}}}\right)}\) Average speed, \(V_{a v g}=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}\)
UP CPMT-2008
Motion in One Dimensions
141333
A boat is sent across a river with a velocity of 8 \(\mathbf{k m} / \mathrm{h}\). If the resultant velocity of boat is 10 \(\mathrm{km} / \mathrm{h}\), then velocity of the river is
1 \(10 \mathrm{~km} / \mathrm{h}\)
2 \(8 \mathrm{~km} / \mathrm{h}\)
3 \(6 \mathrm{~km} / \mathrm{h}\)
4 \(4 \mathrm{~km} / \mathrm{h}\)
Explanation:
C Velocity of boat \(\mathrm{v}_{\mathrm{b}}=8 \mathrm{~km} / \mathrm{hr}\) Resultant Velocity \(\left(\mathrm{v}_{\mathrm{r}}\right)=10 \mathrm{~km} / \mathrm{hr}\) \(\mathrm{v}_{\mathrm{r}}^{2}=\mathrm{u}^{2}+\mathrm{v}_{\mathrm{b}}^{2}\) Velocity of river \((u)=\sqrt{v_{r}^{2}-v_{b}^{2}}\) \(=\sqrt{(10)^{2}-(8)^{2}}=6 \mathrm{~km} / \mathrm{hr}\)
UP CPMT-2010
Motion in One Dimensions
141334
A point initially at rest moves along \(x\)-axis. Its acceleration varies with time as \(a=(6 t+5)\) \(\mathrm{m} / \mathrm{s}^{2}\). If it starts from origin, the distance covered in \(2 \mathrm{~s}\) is
1 \(20 \mathrm{~m}\)
2 \(18 \mathrm{~m}\)
3 \(16 \mathrm{~m}\)
4 \(25 \mathrm{~m}\)
Explanation:
B Given, the acceleration of the particle varies with time as; \(a=(6 t+5) \mathrm{m} / \mathrm{s}^{2}\) Integrating the equation (i), we get- \(\int(6 t+5) d t=v\) \(\frac{6 t^{2}}{2}+5 t=v\) \(3 t^{2}+5 t=v\) Now, given that the velocity of the particle varies with time as \(v=\left(3 t^{2}+5 t\right) \mathrm{ms}^{-1}\). Integrating the equation, \(\int\left(3 \mathrm{t}^{2}+5 \mathrm{t}\right) \mathrm{dt}=\mathrm{x}\) \(\frac{3 \mathrm{t}^{3}}{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x} \Rightarrow \mathrm{t}^{3}+\frac{5 \mathrm{t}^{2}}{2}=\mathrm{x}\) At, \(\quad \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{x}=2^{3}+\frac{5 \times 2^{2}}{2}=18 \mathrm{~m}\)
UP CPMT-2006
Motion in One Dimensions
141335
If relation between distance and time is \(s=a+b t+c t^{2}\), find initial velocity and acceleration.
1 \(b+2 c t, 2 c\)
2 b, 2c
3 \(2 \mathrm{c}, \mathrm{b}\)
4 \(b+2 c, 2 c\)
Explanation:
B Given, if relation between distance and time is \(s=a+b t+c t^{2}\) Differentiation both sides w.r.t time, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0+\mathrm{b}+2 \mathrm{ct}=\mathrm{b}+2 \mathrm{ct}\) Initial velocity \(=\left|\frac{\mathrm{ds}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=\mathrm{b}+2 \mathrm{c} \times 0\) \(=\mathrm{b}+0=\mathrm{b}\) \(\because \quad\) Acceleration \((a)=\frac{d v}{d t}=2 c\) Initial acceleration \(\left|\frac{\mathrm{dv}}{\mathrm{dt}}\right|_{\mathrm{t}=0}=2 \mathrm{c}\) Hence, initial velocity \(=\mathrm{b}\) and initial acceleration \(=2 \mathrm{c}\)
UP CPMT-2003
Motion in One Dimensions
141336
A ball is thrown upwards, it takes \(4 \mathrm{~s}\) to reach back to the ground. Find its initial velocity
1 \(30 \mathrm{~ms}^{-1}\)
2 \(10 \mathrm{~ms}^{-1}\)
3 \(40 \mathrm{~ms}^{-1}\)
4 \(20 \mathrm{~ms}^{-1}\)
Explanation:
D Given that, Time \((\mathrm{t})=4 \mathrm{sec}(\) for whole trip \()\) \(\therefore\) Time to reach maximum height is half of total time \(\mathrm{t}_{\max }=\frac{4}{2}=2 \mathrm{sec}\) Also speed at maximum height is zero, \(\therefore \quad\) Final velocity \((\mathrm{v})=0\) We know that, \(\mathrm{v}=\mathrm{u}-\mathrm{g}_{\max }\) \(\mathrm{u}=0+10 \times 2=20 \mathrm{~m} / \mathrm{s}\)
