141696
A particle moves along a straight line \(O X\). At a time \(t\) (in second), the distance \(x\) (in metre) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\). How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given, \(x=40+12 t-t^{3}\) Velocity \((\mathrm{v})=\frac{\mathrm{dx}}{\mathrm{dt}}=0+12-3 \mathrm{t}^{2}\) At coming to rest \((\mathrm{v}=0)\), \(0=12-3 \mathrm{t}^{2}\) \(\mathrm{t}^{2}=4 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) The distance travelled by particle before coming to rest \(x=40+12(2)-(2)^{3}\) \(x=56 \mathrm{~m}\)
AIPMT 2006
Motion in One Dimensions
141697
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\), in the third second is.
1 \(10 / 3 \mathrm{~m}\)
2 \(19 / 3 \mathrm{~m}\)
3 \(6 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
A Given, Acceleration \((a)=\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\) Distance covered in \(\mathrm{n}^{\text {th }}\) second \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)\) Distance covered by particle in third second is- \(\mathrm{s}_{3}=0+\frac{1}{2} \times \frac{4}{3}(2 \times 3-1) \quad(\text { initial velocity } \mathrm{u}=0)\) \(\mathrm{s}_{3}=\frac{10}{3} \mathrm{~m}\)
CG PET-22.05.2022
Motion in One Dimensions
141698
A ball under uniform acceleration travels \(6 \mathrm{~m}\) in first \(2 \mathrm{~s}\) and \(16 \mathrm{~m}\) in the next \(2 \mathrm{~s}\). Its initial velocity is
1 \(\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
2 \(1 \mathrm{~m} / \mathrm{s}\)
3 \(\frac{8}{3} \mathrm{~m} / \mathrm{s}\)
4 \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\)
Explanation:
A Given, Distance travelled \(\left(\mathrm{s}_{1}\right)=6 \mathrm{~m}\) Distance travelled \(\left(\mathrm{s}_{2}\right)=16 \mathrm{~m}\) Time taken by \(\mathrm{s}_{1}\left(\mathrm{t}_{1}\right)=2 \mathrm{~s}\) Time taken by \(\mathrm{s}_{2}\left(\mathrm{t}_{2}\right)=4 \mathrm{~s}\) Let the initial velocity be \(u\) We know, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) When \(\mathrm{s}_{1}=6 \mathrm{~m}, \mathrm{t}_{1}=2 \mathrm{~s}\) \(6=2 \mathrm{u}+2 \mathrm{a}\) and when \(\mathrm{s}=\mathrm{s}_{1}+\mathrm{s}_{2}=22 \mathrm{~m}\) \(\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=4 \mathrm{~s}\) \(22=4 u+8 a\) On solving equation (1) and equation (2), we get \(\mathrm{u}=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141699
The position of a particle is \(r=x i+y j\) where \(x\) and \(y\) are function of time \(t\) and given as \(x=\left(12+5 t-t^{2}\right) m\) and \(y=\left(18+5 t-t^{2}\right) m\). At \(t=\) \(1 \mathrm{~s}\), the magnitude of the velocity vector of the particle is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
2 \(3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
3 \(4 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, Position \((\overrightarrow{\mathrm{r}})=\mathrm{xi}+\mathrm{yj}\) Where, \(x=12+5 t-t^{2}\) \(\mathrm{y}=18+5 \mathrm{t}-\mathrm{t}^{2}\) We know, \(\overrightarrow{\mathrm{v}}=\frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}\) Differentiating the \(\overrightarrow{\mathrm{r}}\), \(\frac{\mathrm{dr}}{\mathrm{dt}}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(|\mathrm{v}|=\sqrt{3^{2}+3^{2}}\) \(|\mathrm{v}|=3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141700
The velocity-time(v-t) relation of a particle moving in a plane is \(v=3 t^{2} \mathrm{~m} / \mathrm{s}\). At \(\mathbf{t}=\mathbf{0}\); displacement \(x=8 \mathrm{~m}\). The velocity of the particle at \(x=16 \mathrm{~m}\) is
1 \(12 \mathrm{~m} / \mathrm{s}\)
2 \(14 \mathrm{~m} / \mathrm{s}\)
3 \(18 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
A Give that, \(\mathrm{v}=3 \mathrm{t}^{2}\) Differentiation of displacement gives velocity i.e. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\) \(\mathrm{vdt}=\mathrm{dx}\) Integrating both side, we get \(\int 3 t^{2} d t=\int d x\) \(\frac{3 t^{3}}{3}+c=x\) \(t^{3}+c=x\) At \(\mathrm{t}=0, \mathrm{x}=8\) \((0)^{3}+c=8 \Rightarrow c=8\) Putting value of \(\mathrm{c}\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\), we get \(\mathrm{t}^{3}+8=\mathrm{x}\) At \(\quad \mathrm{x}=16 \mathrm{~m}, \mathrm{t}=\) ? \(\mathrm{t}^{3}+8=16 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) Then, \(\quad v=3 t^{2}\) \(=3 \times(2)^{2}\) \(=12 \mathrm{~m} / \mathrm{s}\)
141696
A particle moves along a straight line \(O X\). At a time \(t\) (in second), the distance \(x\) (in metre) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\). How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given, \(x=40+12 t-t^{3}\) Velocity \((\mathrm{v})=\frac{\mathrm{dx}}{\mathrm{dt}}=0+12-3 \mathrm{t}^{2}\) At coming to rest \((\mathrm{v}=0)\), \(0=12-3 \mathrm{t}^{2}\) \(\mathrm{t}^{2}=4 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) The distance travelled by particle before coming to rest \(x=40+12(2)-(2)^{3}\) \(x=56 \mathrm{~m}\)
AIPMT 2006
Motion in One Dimensions
141697
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\), in the third second is.
1 \(10 / 3 \mathrm{~m}\)
2 \(19 / 3 \mathrm{~m}\)
3 \(6 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
A Given, Acceleration \((a)=\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\) Distance covered in \(\mathrm{n}^{\text {th }}\) second \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)\) Distance covered by particle in third second is- \(\mathrm{s}_{3}=0+\frac{1}{2} \times \frac{4}{3}(2 \times 3-1) \quad(\text { initial velocity } \mathrm{u}=0)\) \(\mathrm{s}_{3}=\frac{10}{3} \mathrm{~m}\)
CG PET-22.05.2022
Motion in One Dimensions
141698
A ball under uniform acceleration travels \(6 \mathrm{~m}\) in first \(2 \mathrm{~s}\) and \(16 \mathrm{~m}\) in the next \(2 \mathrm{~s}\). Its initial velocity is
1 \(\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
2 \(1 \mathrm{~m} / \mathrm{s}\)
3 \(\frac{8}{3} \mathrm{~m} / \mathrm{s}\)
4 \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\)
Explanation:
A Given, Distance travelled \(\left(\mathrm{s}_{1}\right)=6 \mathrm{~m}\) Distance travelled \(\left(\mathrm{s}_{2}\right)=16 \mathrm{~m}\) Time taken by \(\mathrm{s}_{1}\left(\mathrm{t}_{1}\right)=2 \mathrm{~s}\) Time taken by \(\mathrm{s}_{2}\left(\mathrm{t}_{2}\right)=4 \mathrm{~s}\) Let the initial velocity be \(u\) We know, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) When \(\mathrm{s}_{1}=6 \mathrm{~m}, \mathrm{t}_{1}=2 \mathrm{~s}\) \(6=2 \mathrm{u}+2 \mathrm{a}\) and when \(\mathrm{s}=\mathrm{s}_{1}+\mathrm{s}_{2}=22 \mathrm{~m}\) \(\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=4 \mathrm{~s}\) \(22=4 u+8 a\) On solving equation (1) and equation (2), we get \(\mathrm{u}=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141699
The position of a particle is \(r=x i+y j\) where \(x\) and \(y\) are function of time \(t\) and given as \(x=\left(12+5 t-t^{2}\right) m\) and \(y=\left(18+5 t-t^{2}\right) m\). At \(t=\) \(1 \mathrm{~s}\), the magnitude of the velocity vector of the particle is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
2 \(3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
3 \(4 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, Position \((\overrightarrow{\mathrm{r}})=\mathrm{xi}+\mathrm{yj}\) Where, \(x=12+5 t-t^{2}\) \(\mathrm{y}=18+5 \mathrm{t}-\mathrm{t}^{2}\) We know, \(\overrightarrow{\mathrm{v}}=\frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}\) Differentiating the \(\overrightarrow{\mathrm{r}}\), \(\frac{\mathrm{dr}}{\mathrm{dt}}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(|\mathrm{v}|=\sqrt{3^{2}+3^{2}}\) \(|\mathrm{v}|=3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141700
The velocity-time(v-t) relation of a particle moving in a plane is \(v=3 t^{2} \mathrm{~m} / \mathrm{s}\). At \(\mathbf{t}=\mathbf{0}\); displacement \(x=8 \mathrm{~m}\). The velocity of the particle at \(x=16 \mathrm{~m}\) is
1 \(12 \mathrm{~m} / \mathrm{s}\)
2 \(14 \mathrm{~m} / \mathrm{s}\)
3 \(18 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
A Give that, \(\mathrm{v}=3 \mathrm{t}^{2}\) Differentiation of displacement gives velocity i.e. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\) \(\mathrm{vdt}=\mathrm{dx}\) Integrating both side, we get \(\int 3 t^{2} d t=\int d x\) \(\frac{3 t^{3}}{3}+c=x\) \(t^{3}+c=x\) At \(\mathrm{t}=0, \mathrm{x}=8\) \((0)^{3}+c=8 \Rightarrow c=8\) Putting value of \(\mathrm{c}\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\), we get \(\mathrm{t}^{3}+8=\mathrm{x}\) At \(\quad \mathrm{x}=16 \mathrm{~m}, \mathrm{t}=\) ? \(\mathrm{t}^{3}+8=16 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) Then, \(\quad v=3 t^{2}\) \(=3 \times(2)^{2}\) \(=12 \mathrm{~m} / \mathrm{s}\)
141696
A particle moves along a straight line \(O X\). At a time \(t\) (in second), the distance \(x\) (in metre) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\). How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given, \(x=40+12 t-t^{3}\) Velocity \((\mathrm{v})=\frac{\mathrm{dx}}{\mathrm{dt}}=0+12-3 \mathrm{t}^{2}\) At coming to rest \((\mathrm{v}=0)\), \(0=12-3 \mathrm{t}^{2}\) \(\mathrm{t}^{2}=4 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) The distance travelled by particle before coming to rest \(x=40+12(2)-(2)^{3}\) \(x=56 \mathrm{~m}\)
AIPMT 2006
Motion in One Dimensions
141697
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\), in the third second is.
1 \(10 / 3 \mathrm{~m}\)
2 \(19 / 3 \mathrm{~m}\)
3 \(6 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
A Given, Acceleration \((a)=\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\) Distance covered in \(\mathrm{n}^{\text {th }}\) second \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)\) Distance covered by particle in third second is- \(\mathrm{s}_{3}=0+\frac{1}{2} \times \frac{4}{3}(2 \times 3-1) \quad(\text { initial velocity } \mathrm{u}=0)\) \(\mathrm{s}_{3}=\frac{10}{3} \mathrm{~m}\)
CG PET-22.05.2022
Motion in One Dimensions
141698
A ball under uniform acceleration travels \(6 \mathrm{~m}\) in first \(2 \mathrm{~s}\) and \(16 \mathrm{~m}\) in the next \(2 \mathrm{~s}\). Its initial velocity is
1 \(\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
2 \(1 \mathrm{~m} / \mathrm{s}\)
3 \(\frac{8}{3} \mathrm{~m} / \mathrm{s}\)
4 \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\)
Explanation:
A Given, Distance travelled \(\left(\mathrm{s}_{1}\right)=6 \mathrm{~m}\) Distance travelled \(\left(\mathrm{s}_{2}\right)=16 \mathrm{~m}\) Time taken by \(\mathrm{s}_{1}\left(\mathrm{t}_{1}\right)=2 \mathrm{~s}\) Time taken by \(\mathrm{s}_{2}\left(\mathrm{t}_{2}\right)=4 \mathrm{~s}\) Let the initial velocity be \(u\) We know, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) When \(\mathrm{s}_{1}=6 \mathrm{~m}, \mathrm{t}_{1}=2 \mathrm{~s}\) \(6=2 \mathrm{u}+2 \mathrm{a}\) and when \(\mathrm{s}=\mathrm{s}_{1}+\mathrm{s}_{2}=22 \mathrm{~m}\) \(\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=4 \mathrm{~s}\) \(22=4 u+8 a\) On solving equation (1) and equation (2), we get \(\mathrm{u}=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141699
The position of a particle is \(r=x i+y j\) where \(x\) and \(y\) are function of time \(t\) and given as \(x=\left(12+5 t-t^{2}\right) m\) and \(y=\left(18+5 t-t^{2}\right) m\). At \(t=\) \(1 \mathrm{~s}\), the magnitude of the velocity vector of the particle is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
2 \(3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
3 \(4 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, Position \((\overrightarrow{\mathrm{r}})=\mathrm{xi}+\mathrm{yj}\) Where, \(x=12+5 t-t^{2}\) \(\mathrm{y}=18+5 \mathrm{t}-\mathrm{t}^{2}\) We know, \(\overrightarrow{\mathrm{v}}=\frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}\) Differentiating the \(\overrightarrow{\mathrm{r}}\), \(\frac{\mathrm{dr}}{\mathrm{dt}}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(|\mathrm{v}|=\sqrt{3^{2}+3^{2}}\) \(|\mathrm{v}|=3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141700
The velocity-time(v-t) relation of a particle moving in a plane is \(v=3 t^{2} \mathrm{~m} / \mathrm{s}\). At \(\mathbf{t}=\mathbf{0}\); displacement \(x=8 \mathrm{~m}\). The velocity of the particle at \(x=16 \mathrm{~m}\) is
1 \(12 \mathrm{~m} / \mathrm{s}\)
2 \(14 \mathrm{~m} / \mathrm{s}\)
3 \(18 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
A Give that, \(\mathrm{v}=3 \mathrm{t}^{2}\) Differentiation of displacement gives velocity i.e. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\) \(\mathrm{vdt}=\mathrm{dx}\) Integrating both side, we get \(\int 3 t^{2} d t=\int d x\) \(\frac{3 t^{3}}{3}+c=x\) \(t^{3}+c=x\) At \(\mathrm{t}=0, \mathrm{x}=8\) \((0)^{3}+c=8 \Rightarrow c=8\) Putting value of \(\mathrm{c}\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\), we get \(\mathrm{t}^{3}+8=\mathrm{x}\) At \(\quad \mathrm{x}=16 \mathrm{~m}, \mathrm{t}=\) ? \(\mathrm{t}^{3}+8=16 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) Then, \(\quad v=3 t^{2}\) \(=3 \times(2)^{2}\) \(=12 \mathrm{~m} / \mathrm{s}\)
141696
A particle moves along a straight line \(O X\). At a time \(t\) (in second), the distance \(x\) (in metre) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\). How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given, \(x=40+12 t-t^{3}\) Velocity \((\mathrm{v})=\frac{\mathrm{dx}}{\mathrm{dt}}=0+12-3 \mathrm{t}^{2}\) At coming to rest \((\mathrm{v}=0)\), \(0=12-3 \mathrm{t}^{2}\) \(\mathrm{t}^{2}=4 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) The distance travelled by particle before coming to rest \(x=40+12(2)-(2)^{3}\) \(x=56 \mathrm{~m}\)
AIPMT 2006
Motion in One Dimensions
141697
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\), in the third second is.
1 \(10 / 3 \mathrm{~m}\)
2 \(19 / 3 \mathrm{~m}\)
3 \(6 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
A Given, Acceleration \((a)=\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\) Distance covered in \(\mathrm{n}^{\text {th }}\) second \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)\) Distance covered by particle in third second is- \(\mathrm{s}_{3}=0+\frac{1}{2} \times \frac{4}{3}(2 \times 3-1) \quad(\text { initial velocity } \mathrm{u}=0)\) \(\mathrm{s}_{3}=\frac{10}{3} \mathrm{~m}\)
CG PET-22.05.2022
Motion in One Dimensions
141698
A ball under uniform acceleration travels \(6 \mathrm{~m}\) in first \(2 \mathrm{~s}\) and \(16 \mathrm{~m}\) in the next \(2 \mathrm{~s}\). Its initial velocity is
1 \(\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
2 \(1 \mathrm{~m} / \mathrm{s}\)
3 \(\frac{8}{3} \mathrm{~m} / \mathrm{s}\)
4 \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\)
Explanation:
A Given, Distance travelled \(\left(\mathrm{s}_{1}\right)=6 \mathrm{~m}\) Distance travelled \(\left(\mathrm{s}_{2}\right)=16 \mathrm{~m}\) Time taken by \(\mathrm{s}_{1}\left(\mathrm{t}_{1}\right)=2 \mathrm{~s}\) Time taken by \(\mathrm{s}_{2}\left(\mathrm{t}_{2}\right)=4 \mathrm{~s}\) Let the initial velocity be \(u\) We know, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) When \(\mathrm{s}_{1}=6 \mathrm{~m}, \mathrm{t}_{1}=2 \mathrm{~s}\) \(6=2 \mathrm{u}+2 \mathrm{a}\) and when \(\mathrm{s}=\mathrm{s}_{1}+\mathrm{s}_{2}=22 \mathrm{~m}\) \(\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=4 \mathrm{~s}\) \(22=4 u+8 a\) On solving equation (1) and equation (2), we get \(\mathrm{u}=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141699
The position of a particle is \(r=x i+y j\) where \(x\) and \(y\) are function of time \(t\) and given as \(x=\left(12+5 t-t^{2}\right) m\) and \(y=\left(18+5 t-t^{2}\right) m\). At \(t=\) \(1 \mathrm{~s}\), the magnitude of the velocity vector of the particle is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
2 \(3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
3 \(4 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, Position \((\overrightarrow{\mathrm{r}})=\mathrm{xi}+\mathrm{yj}\) Where, \(x=12+5 t-t^{2}\) \(\mathrm{y}=18+5 \mathrm{t}-\mathrm{t}^{2}\) We know, \(\overrightarrow{\mathrm{v}}=\frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}\) Differentiating the \(\overrightarrow{\mathrm{r}}\), \(\frac{\mathrm{dr}}{\mathrm{dt}}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(|\mathrm{v}|=\sqrt{3^{2}+3^{2}}\) \(|\mathrm{v}|=3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141700
The velocity-time(v-t) relation of a particle moving in a plane is \(v=3 t^{2} \mathrm{~m} / \mathrm{s}\). At \(\mathbf{t}=\mathbf{0}\); displacement \(x=8 \mathrm{~m}\). The velocity of the particle at \(x=16 \mathrm{~m}\) is
1 \(12 \mathrm{~m} / \mathrm{s}\)
2 \(14 \mathrm{~m} / \mathrm{s}\)
3 \(18 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
A Give that, \(\mathrm{v}=3 \mathrm{t}^{2}\) Differentiation of displacement gives velocity i.e. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\) \(\mathrm{vdt}=\mathrm{dx}\) Integrating both side, we get \(\int 3 t^{2} d t=\int d x\) \(\frac{3 t^{3}}{3}+c=x\) \(t^{3}+c=x\) At \(\mathrm{t}=0, \mathrm{x}=8\) \((0)^{3}+c=8 \Rightarrow c=8\) Putting value of \(\mathrm{c}\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\), we get \(\mathrm{t}^{3}+8=\mathrm{x}\) At \(\quad \mathrm{x}=16 \mathrm{~m}, \mathrm{t}=\) ? \(\mathrm{t}^{3}+8=16 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) Then, \(\quad v=3 t^{2}\) \(=3 \times(2)^{2}\) \(=12 \mathrm{~m} / \mathrm{s}\)
141696
A particle moves along a straight line \(O X\). At a time \(t\) (in second), the distance \(x\) (in metre) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\). How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given, \(x=40+12 t-t^{3}\) Velocity \((\mathrm{v})=\frac{\mathrm{dx}}{\mathrm{dt}}=0+12-3 \mathrm{t}^{2}\) At coming to rest \((\mathrm{v}=0)\), \(0=12-3 \mathrm{t}^{2}\) \(\mathrm{t}^{2}=4 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) The distance travelled by particle before coming to rest \(x=40+12(2)-(2)^{3}\) \(x=56 \mathrm{~m}\)
AIPMT 2006
Motion in One Dimensions
141697
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\), in the third second is.
1 \(10 / 3 \mathrm{~m}\)
2 \(19 / 3 \mathrm{~m}\)
3 \(6 \mathrm{~m}\)
4 \(4 \mathrm{~m}\)
Explanation:
A Given, Acceleration \((a)=\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\) Distance covered in \(\mathrm{n}^{\text {th }}\) second \(\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)\) Distance covered by particle in third second is- \(\mathrm{s}_{3}=0+\frac{1}{2} \times \frac{4}{3}(2 \times 3-1) \quad(\text { initial velocity } \mathrm{u}=0)\) \(\mathrm{s}_{3}=\frac{10}{3} \mathrm{~m}\)
CG PET-22.05.2022
Motion in One Dimensions
141698
A ball under uniform acceleration travels \(6 \mathrm{~m}\) in first \(2 \mathrm{~s}\) and \(16 \mathrm{~m}\) in the next \(2 \mathrm{~s}\). Its initial velocity is
1 \(\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
2 \(1 \mathrm{~m} / \mathrm{s}\)
3 \(\frac{8}{3} \mathrm{~m} / \mathrm{s}\)
4 \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\)
Explanation:
A Given, Distance travelled \(\left(\mathrm{s}_{1}\right)=6 \mathrm{~m}\) Distance travelled \(\left(\mathrm{s}_{2}\right)=16 \mathrm{~m}\) Time taken by \(\mathrm{s}_{1}\left(\mathrm{t}_{1}\right)=2 \mathrm{~s}\) Time taken by \(\mathrm{s}_{2}\left(\mathrm{t}_{2}\right)=4 \mathrm{~s}\) Let the initial velocity be \(u\) We know, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) When \(\mathrm{s}_{1}=6 \mathrm{~m}, \mathrm{t}_{1}=2 \mathrm{~s}\) \(6=2 \mathrm{u}+2 \mathrm{a}\) and when \(\mathrm{s}=\mathrm{s}_{1}+\mathrm{s}_{2}=22 \mathrm{~m}\) \(\mathrm{t}=\mathrm{t}_{1}+\mathrm{t}_{2}=4 \mathrm{~s}\) \(22=4 u+8 a\) On solving equation (1) and equation (2), we get \(\mathrm{u}=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141699
The position of a particle is \(r=x i+y j\) where \(x\) and \(y\) are function of time \(t\) and given as \(x=\left(12+5 t-t^{2}\right) m\) and \(y=\left(18+5 t-t^{2}\right) m\). At \(t=\) \(1 \mathrm{~s}\), the magnitude of the velocity vector of the particle is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
2 \(3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
3 \(4 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, Position \((\overrightarrow{\mathrm{r}})=\mathrm{xi}+\mathrm{yj}\) Where, \(x=12+5 t-t^{2}\) \(\mathrm{y}=18+5 \mathrm{t}-\mathrm{t}^{2}\) We know, \(\overrightarrow{\mathrm{v}}=\frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}\) Differentiating the \(\overrightarrow{\mathrm{r}}\), \(\frac{\mathrm{dr}}{\mathrm{dt}}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=(5-2 \mathrm{t}) \hat{\mathrm{i}}+(5-2 \mathrm{t}) \hat{\mathrm{j}}\) \(\mathrm{v}_{\mathrm{t}}=1 \mathrm{~s}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(|\mathrm{v}|=\sqrt{3^{2}+3^{2}}\) \(|\mathrm{v}|=3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
TS EAMCET 30.07.2022
Motion in One Dimensions
141700
The velocity-time(v-t) relation of a particle moving in a plane is \(v=3 t^{2} \mathrm{~m} / \mathrm{s}\). At \(\mathbf{t}=\mathbf{0}\); displacement \(x=8 \mathrm{~m}\). The velocity of the particle at \(x=16 \mathrm{~m}\) is
1 \(12 \mathrm{~m} / \mathrm{s}\)
2 \(14 \mathrm{~m} / \mathrm{s}\)
3 \(18 \mathrm{~m} / \mathrm{s}\)
4 \(10 \mathrm{~m} / \mathrm{s}\)
Explanation:
A Give that, \(\mathrm{v}=3 \mathrm{t}^{2}\) Differentiation of displacement gives velocity i.e. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\) \(\mathrm{vdt}=\mathrm{dx}\) Integrating both side, we get \(\int 3 t^{2} d t=\int d x\) \(\frac{3 t^{3}}{3}+c=x\) \(t^{3}+c=x\) At \(\mathrm{t}=0, \mathrm{x}=8\) \((0)^{3}+c=8 \Rightarrow c=8\) Putting value of \(\mathrm{c}\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\), we get \(\mathrm{t}^{3}+8=\mathrm{x}\) At \(\quad \mathrm{x}=16 \mathrm{~m}, \mathrm{t}=\) ? \(\mathrm{t}^{3}+8=16 \Rightarrow \mathrm{t}=2 \mathrm{sec}\) Then, \(\quad v=3 t^{2}\) \(=3 \times(2)^{2}\) \(=12 \mathrm{~m} / \mathrm{s}\)
