141193
The velocity- displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by
1
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4
Explanation:
C Given, initial velocity \(=\mathrm{v}_{0}\) Let, the distance travelled in time \(t=x_{0}\) For the graph \(\tan \theta=\frac{-\mathrm{v}_{0}}{\mathrm{x}_{0}}\) Therefore \(\mathrm{v}=\left(-\frac{\mathrm{v}_{0}}{\mathrm{x}_{0}}\right) \mathrm{x}+\mathrm{v}_{0}\) And, \(\quad a=\frac{v d v}{d x}\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right] \cdot \frac{d}{d x}\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\left[\left(-\frac{v_{0}}{x_{0}}\right)+0\right]\) \(a=\frac{v_{0}^{2}}{x_{0}^{2}} x-\frac{v_{0}^{2}}{x_{0}}\) Again comparing with standard equation of straight line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) here, \(m=+v e, \quad c=-v e\)
Shift-II]
Motion in One Dimensions
141194
A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) to come to rest. If the total time elapsed is \(t\) seconds, the total distance travelled is
C The slope of line OP, \(\alpha=\frac{v_{\max }}{t_{1}} \Rightarrow v_{\max }=\alpha t_{1}\) Slope of line PQ \(\beta=\frac{v_{\max }}{t-t_{1}} \Rightarrow v_{\max }=\beta\left(t-t_{1}\right)\) From equation (i) and (ii) we get \(\alpha \mathrm{t}_{1}=\beta\left(\mathrm{t}-\mathrm{t}_{1}\right)\) \(\alpha \mathrm{t}_{1}=\beta \mathrm{t}-\beta \mathrm{t}_{1}\) \(\alpha \mathrm{t}_{1}+\beta \mathrm{t}_{1}=\beta \mathrm{t}\) \(\mathrm{t}_{1}(\alpha+\beta)=\beta \mathrm{t}\) \(\mathrm{t}_{1}=\frac{\beta \mathrm{t}}{\alpha+\beta} \cdots\) \(\because\) Total distance travelled \(\mathrm{s}=\) Area of \(\mathrm{v}-\mathrm{t}\) graph \(=\frac{1}{2} \times \mathrm{v}_{\max } \times \mathrm{t}\) \(\mathrm{s}=\frac{1}{2} \mathrm{v}_{\max } \mathrm{t}\) Substituting value of \(t_{1}\) in equation (i) we get \(\mathrm{v}_{\max }=\frac{\alpha \beta \mathrm{t}}{\alpha+\beta}\) Substituting value of \(\mathrm{v}_{\max }\) in equation (iv) we get \(\mathrm{s}=\frac{\alpha \beta}{2(\alpha+\beta)} \mathrm{t}^{2}\)
AIPMT 1994
Motion in One Dimensions
141195
The velocity-displacement graph describing the motion of a bicycle is shown in the following figure. The acceleration-displacement graph of the bicycle's motion is best described by.
1
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4
Explanation:
A From \(\mathrm{x}=0\) to \(200 \mathrm{~m}\), \(\because \mathrm{y}=\mathrm{mx}+\mathrm{c}\) \(\mathrm{v}=\mathrm{mx}+\mathrm{c}\) \(=\frac{40}{200} \mathrm{x}+\mathrm{c}\) \(\mathrm{v}=\frac{1}{5} \mathrm{x}+\mathrm{c}\) \(a=\frac{v d v}{d x}\) \(a=\left(\frac{x}{5}+10\right)\left(\frac{1}{5}\right)\) \(a=\frac{x}{25}+2\) Straight line till \(x=200\) \(\because\) For \(x>200, v=\) constant, \(a=0\) \(a\left(\mathrm{~ms}^{2}\right)\)
Shift-I]
Motion in One Dimensions
141196
A scooter accelerates from rest for time \(t_{1}\) at constant rate \(a_{1}\) and then retards at constant rate \(a_{2}\) for time \(t_{2}\) and comes to rest. The correct value of \(\frac{t_{1}}{t_{2}}\) will be
1 \(\frac{a_{1}+a_{2}}{a_{2}}\)
2 \(\frac{a_{2}}{a_{1}}\)
3 \(\frac{a_{1}}{a_{2}}\)
4 \(\frac{a_{1}+a_{2}}{a_{1}}\)
Explanation:
B From (v-t) graph \(\tan \theta_{1}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{1}}=\mathrm{a}_{1}\) \(\tan \theta_{2}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{2}}=\mathrm{a}_{2}\) From equation (i) and (ii) we get \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{\max } / \mathrm{t}_{1}}{\mathrm{v}_{\max } / \mathrm{t}_{2}}=\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) \(\because \quad \frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}\)
141193
The velocity- displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by
1
2
3
4
Explanation:
C Given, initial velocity \(=\mathrm{v}_{0}\) Let, the distance travelled in time \(t=x_{0}\) For the graph \(\tan \theta=\frac{-\mathrm{v}_{0}}{\mathrm{x}_{0}}\) Therefore \(\mathrm{v}=\left(-\frac{\mathrm{v}_{0}}{\mathrm{x}_{0}}\right) \mathrm{x}+\mathrm{v}_{0}\) And, \(\quad a=\frac{v d v}{d x}\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right] \cdot \frac{d}{d x}\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\left[\left(-\frac{v_{0}}{x_{0}}\right)+0\right]\) \(a=\frac{v_{0}^{2}}{x_{0}^{2}} x-\frac{v_{0}^{2}}{x_{0}}\) Again comparing with standard equation of straight line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) here, \(m=+v e, \quad c=-v e\)
Shift-II]
Motion in One Dimensions
141194
A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) to come to rest. If the total time elapsed is \(t\) seconds, the total distance travelled is
C The slope of line OP, \(\alpha=\frac{v_{\max }}{t_{1}} \Rightarrow v_{\max }=\alpha t_{1}\) Slope of line PQ \(\beta=\frac{v_{\max }}{t-t_{1}} \Rightarrow v_{\max }=\beta\left(t-t_{1}\right)\) From equation (i) and (ii) we get \(\alpha \mathrm{t}_{1}=\beta\left(\mathrm{t}-\mathrm{t}_{1}\right)\) \(\alpha \mathrm{t}_{1}=\beta \mathrm{t}-\beta \mathrm{t}_{1}\) \(\alpha \mathrm{t}_{1}+\beta \mathrm{t}_{1}=\beta \mathrm{t}\) \(\mathrm{t}_{1}(\alpha+\beta)=\beta \mathrm{t}\) \(\mathrm{t}_{1}=\frac{\beta \mathrm{t}}{\alpha+\beta} \cdots\) \(\because\) Total distance travelled \(\mathrm{s}=\) Area of \(\mathrm{v}-\mathrm{t}\) graph \(=\frac{1}{2} \times \mathrm{v}_{\max } \times \mathrm{t}\) \(\mathrm{s}=\frac{1}{2} \mathrm{v}_{\max } \mathrm{t}\) Substituting value of \(t_{1}\) in equation (i) we get \(\mathrm{v}_{\max }=\frac{\alpha \beta \mathrm{t}}{\alpha+\beta}\) Substituting value of \(\mathrm{v}_{\max }\) in equation (iv) we get \(\mathrm{s}=\frac{\alpha \beta}{2(\alpha+\beta)} \mathrm{t}^{2}\)
AIPMT 1994
Motion in One Dimensions
141195
The velocity-displacement graph describing the motion of a bicycle is shown in the following figure. The acceleration-displacement graph of the bicycle's motion is best described by.
1
2
3
4
Explanation:
A From \(\mathrm{x}=0\) to \(200 \mathrm{~m}\), \(\because \mathrm{y}=\mathrm{mx}+\mathrm{c}\) \(\mathrm{v}=\mathrm{mx}+\mathrm{c}\) \(=\frac{40}{200} \mathrm{x}+\mathrm{c}\) \(\mathrm{v}=\frac{1}{5} \mathrm{x}+\mathrm{c}\) \(a=\frac{v d v}{d x}\) \(a=\left(\frac{x}{5}+10\right)\left(\frac{1}{5}\right)\) \(a=\frac{x}{25}+2\) Straight line till \(x=200\) \(\because\) For \(x>200, v=\) constant, \(a=0\) \(a\left(\mathrm{~ms}^{2}\right)\)
Shift-I]
Motion in One Dimensions
141196
A scooter accelerates from rest for time \(t_{1}\) at constant rate \(a_{1}\) and then retards at constant rate \(a_{2}\) for time \(t_{2}\) and comes to rest. The correct value of \(\frac{t_{1}}{t_{2}}\) will be
1 \(\frac{a_{1}+a_{2}}{a_{2}}\)
2 \(\frac{a_{2}}{a_{1}}\)
3 \(\frac{a_{1}}{a_{2}}\)
4 \(\frac{a_{1}+a_{2}}{a_{1}}\)
Explanation:
B From (v-t) graph \(\tan \theta_{1}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{1}}=\mathrm{a}_{1}\) \(\tan \theta_{2}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{2}}=\mathrm{a}_{2}\) From equation (i) and (ii) we get \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{\max } / \mathrm{t}_{1}}{\mathrm{v}_{\max } / \mathrm{t}_{2}}=\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) \(\because \quad \frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}\)
141193
The velocity- displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by
1
2
3
4
Explanation:
C Given, initial velocity \(=\mathrm{v}_{0}\) Let, the distance travelled in time \(t=x_{0}\) For the graph \(\tan \theta=\frac{-\mathrm{v}_{0}}{\mathrm{x}_{0}}\) Therefore \(\mathrm{v}=\left(-\frac{\mathrm{v}_{0}}{\mathrm{x}_{0}}\right) \mathrm{x}+\mathrm{v}_{0}\) And, \(\quad a=\frac{v d v}{d x}\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right] \cdot \frac{d}{d x}\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\left[\left(-\frac{v_{0}}{x_{0}}\right)+0\right]\) \(a=\frac{v_{0}^{2}}{x_{0}^{2}} x-\frac{v_{0}^{2}}{x_{0}}\) Again comparing with standard equation of straight line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) here, \(m=+v e, \quad c=-v e\)
Shift-II]
Motion in One Dimensions
141194
A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) to come to rest. If the total time elapsed is \(t\) seconds, the total distance travelled is
C The slope of line OP, \(\alpha=\frac{v_{\max }}{t_{1}} \Rightarrow v_{\max }=\alpha t_{1}\) Slope of line PQ \(\beta=\frac{v_{\max }}{t-t_{1}} \Rightarrow v_{\max }=\beta\left(t-t_{1}\right)\) From equation (i) and (ii) we get \(\alpha \mathrm{t}_{1}=\beta\left(\mathrm{t}-\mathrm{t}_{1}\right)\) \(\alpha \mathrm{t}_{1}=\beta \mathrm{t}-\beta \mathrm{t}_{1}\) \(\alpha \mathrm{t}_{1}+\beta \mathrm{t}_{1}=\beta \mathrm{t}\) \(\mathrm{t}_{1}(\alpha+\beta)=\beta \mathrm{t}\) \(\mathrm{t}_{1}=\frac{\beta \mathrm{t}}{\alpha+\beta} \cdots\) \(\because\) Total distance travelled \(\mathrm{s}=\) Area of \(\mathrm{v}-\mathrm{t}\) graph \(=\frac{1}{2} \times \mathrm{v}_{\max } \times \mathrm{t}\) \(\mathrm{s}=\frac{1}{2} \mathrm{v}_{\max } \mathrm{t}\) Substituting value of \(t_{1}\) in equation (i) we get \(\mathrm{v}_{\max }=\frac{\alpha \beta \mathrm{t}}{\alpha+\beta}\) Substituting value of \(\mathrm{v}_{\max }\) in equation (iv) we get \(\mathrm{s}=\frac{\alpha \beta}{2(\alpha+\beta)} \mathrm{t}^{2}\)
AIPMT 1994
Motion in One Dimensions
141195
The velocity-displacement graph describing the motion of a bicycle is shown in the following figure. The acceleration-displacement graph of the bicycle's motion is best described by.
1
2
3
4
Explanation:
A From \(\mathrm{x}=0\) to \(200 \mathrm{~m}\), \(\because \mathrm{y}=\mathrm{mx}+\mathrm{c}\) \(\mathrm{v}=\mathrm{mx}+\mathrm{c}\) \(=\frac{40}{200} \mathrm{x}+\mathrm{c}\) \(\mathrm{v}=\frac{1}{5} \mathrm{x}+\mathrm{c}\) \(a=\frac{v d v}{d x}\) \(a=\left(\frac{x}{5}+10\right)\left(\frac{1}{5}\right)\) \(a=\frac{x}{25}+2\) Straight line till \(x=200\) \(\because\) For \(x>200, v=\) constant, \(a=0\) \(a\left(\mathrm{~ms}^{2}\right)\)
Shift-I]
Motion in One Dimensions
141196
A scooter accelerates from rest for time \(t_{1}\) at constant rate \(a_{1}\) and then retards at constant rate \(a_{2}\) for time \(t_{2}\) and comes to rest. The correct value of \(\frac{t_{1}}{t_{2}}\) will be
1 \(\frac{a_{1}+a_{2}}{a_{2}}\)
2 \(\frac{a_{2}}{a_{1}}\)
3 \(\frac{a_{1}}{a_{2}}\)
4 \(\frac{a_{1}+a_{2}}{a_{1}}\)
Explanation:
B From (v-t) graph \(\tan \theta_{1}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{1}}=\mathrm{a}_{1}\) \(\tan \theta_{2}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{2}}=\mathrm{a}_{2}\) From equation (i) and (ii) we get \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{\max } / \mathrm{t}_{1}}{\mathrm{v}_{\max } / \mathrm{t}_{2}}=\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) \(\because \quad \frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141193
The velocity- displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by
1
2
3
4
Explanation:
C Given, initial velocity \(=\mathrm{v}_{0}\) Let, the distance travelled in time \(t=x_{0}\) For the graph \(\tan \theta=\frac{-\mathrm{v}_{0}}{\mathrm{x}_{0}}\) Therefore \(\mathrm{v}=\left(-\frac{\mathrm{v}_{0}}{\mathrm{x}_{0}}\right) \mathrm{x}+\mathrm{v}_{0}\) And, \(\quad a=\frac{v d v}{d x}\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right] \cdot \frac{d}{d x}\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\) \(a=\left[\left(-\frac{v_{0}}{x_{0}}\right) x+v_{0}\right]\left[\left(-\frac{v_{0}}{x_{0}}\right)+0\right]\) \(a=\frac{v_{0}^{2}}{x_{0}^{2}} x-\frac{v_{0}^{2}}{x_{0}}\) Again comparing with standard equation of straight line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) here, \(m=+v e, \quad c=-v e\)
Shift-II]
Motion in One Dimensions
141194
A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) to come to rest. If the total time elapsed is \(t\) seconds, the total distance travelled is
C The slope of line OP, \(\alpha=\frac{v_{\max }}{t_{1}} \Rightarrow v_{\max }=\alpha t_{1}\) Slope of line PQ \(\beta=\frac{v_{\max }}{t-t_{1}} \Rightarrow v_{\max }=\beta\left(t-t_{1}\right)\) From equation (i) and (ii) we get \(\alpha \mathrm{t}_{1}=\beta\left(\mathrm{t}-\mathrm{t}_{1}\right)\) \(\alpha \mathrm{t}_{1}=\beta \mathrm{t}-\beta \mathrm{t}_{1}\) \(\alpha \mathrm{t}_{1}+\beta \mathrm{t}_{1}=\beta \mathrm{t}\) \(\mathrm{t}_{1}(\alpha+\beta)=\beta \mathrm{t}\) \(\mathrm{t}_{1}=\frac{\beta \mathrm{t}}{\alpha+\beta} \cdots\) \(\because\) Total distance travelled \(\mathrm{s}=\) Area of \(\mathrm{v}-\mathrm{t}\) graph \(=\frac{1}{2} \times \mathrm{v}_{\max } \times \mathrm{t}\) \(\mathrm{s}=\frac{1}{2} \mathrm{v}_{\max } \mathrm{t}\) Substituting value of \(t_{1}\) in equation (i) we get \(\mathrm{v}_{\max }=\frac{\alpha \beta \mathrm{t}}{\alpha+\beta}\) Substituting value of \(\mathrm{v}_{\max }\) in equation (iv) we get \(\mathrm{s}=\frac{\alpha \beta}{2(\alpha+\beta)} \mathrm{t}^{2}\)
AIPMT 1994
Motion in One Dimensions
141195
The velocity-displacement graph describing the motion of a bicycle is shown in the following figure. The acceleration-displacement graph of the bicycle's motion is best described by.
1
2
3
4
Explanation:
A From \(\mathrm{x}=0\) to \(200 \mathrm{~m}\), \(\because \mathrm{y}=\mathrm{mx}+\mathrm{c}\) \(\mathrm{v}=\mathrm{mx}+\mathrm{c}\) \(=\frac{40}{200} \mathrm{x}+\mathrm{c}\) \(\mathrm{v}=\frac{1}{5} \mathrm{x}+\mathrm{c}\) \(a=\frac{v d v}{d x}\) \(a=\left(\frac{x}{5}+10\right)\left(\frac{1}{5}\right)\) \(a=\frac{x}{25}+2\) Straight line till \(x=200\) \(\because\) For \(x>200, v=\) constant, \(a=0\) \(a\left(\mathrm{~ms}^{2}\right)\)
Shift-I]
Motion in One Dimensions
141196
A scooter accelerates from rest for time \(t_{1}\) at constant rate \(a_{1}\) and then retards at constant rate \(a_{2}\) for time \(t_{2}\) and comes to rest. The correct value of \(\frac{t_{1}}{t_{2}}\) will be
1 \(\frac{a_{1}+a_{2}}{a_{2}}\)
2 \(\frac{a_{2}}{a_{1}}\)
3 \(\frac{a_{1}}{a_{2}}\)
4 \(\frac{a_{1}+a_{2}}{a_{1}}\)
Explanation:
B From (v-t) graph \(\tan \theta_{1}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{1}}=\mathrm{a}_{1}\) \(\tan \theta_{2}=\frac{\mathrm{v}_{\text {max }}}{\mathrm{t}_{2}}=\mathrm{a}_{2}\) From equation (i) and (ii) we get \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{\max } / \mathrm{t}_{1}}{\mathrm{v}_{\max } / \mathrm{t}_{2}}=\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) \(\because \quad \frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}\)
