141267
A particle starts from rest and experiences constant acceleration for 6 seconds. If it travels a distance \(d_{1}\) in the first two seconds, a distance \(d_{2}\) in the next two seconds and a distance \(d_{3}\) in the last two seconds, then
C From the second equation of motion, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here \(\mathrm{u}=0\) So, \(\mathrm{s}=0+\frac{1}{2} \mathrm{at}^{2}=\frac{1}{2} \mathrm{at}^{2}\) Now according to the question, \(d_{1}=\frac{1}{2} a(2)^{2}=2 a\) \(d_{1}+d_{2}=\frac{1}{2} a(4)^{2}=8 a\) \(d_{2}=\left(d_{1}+d_{2}\right)-d_{1}=8 a-2 a=6 a\) \(d_{1}+d_{2}+d_{3}=\frac{1}{2} a(6)^{2}\) \(\because d_{3}=\left(d_{1}+d_{2}+d_{3}\right)-\left(d_{1}+d_{2}\right)\) \(d_{3}=\frac{1}{2} a(6)^{2}-8 a=10 a\) Hence, \(d_{1}: d_{2}: d_{3}=2 a: 6 a: 10 a\) \(=1: 3: 5\)
J and K-CET- 2003
Motion in One Dimensions
141269
A body is travelling east with a speed of \(9 \mathrm{~m} / \mathrm{s}\) and with an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) acting west on it. The displacement of the body during the \(5^{\text {th }}\) second of its motion is
1 \(0.25 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.75 \mathrm{~m}\)
4 zero
Explanation:
D Given, \(\mathrm{u}=9 \mathrm{~m} / \mathrm{s}\) east direction \(\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) west direction Distance covered in \(\mathrm{t}^{\text {th }}\) second (retardation)- \(\mathrm{s}_{\mathrm{t}}=\mathrm{u}-\frac{1}{2} \mathrm{a}(2 \mathrm{t}-1) \quad\left\{\because \mathrm{s}_{\mathrm{t}}=5 \mathrm{sec}\right\}\) \(\mathrm{s}_{5}=9-\frac{1}{2} \times 2(2 \times 5-1)=9-(10-1)\) \(\mathrm{s}_{5}=0\)
J and K-CET-2015
Motion in One Dimensions
141271
For a particle moving according to the equation \(x=a \cos \pi t\), the displacement in \(3 \mathrm{~s}\) is
1 0
2 \(0.5 \mathrm{a}\)
3 \(1.5 \mathrm{a}\)
4 \(2 \mathrm{a}\)
5 \(\mathrm{a}\)
Explanation:
D Given that, At \(\mathrm{t}=0, \mathrm{x}=\mathrm{a} \cos \pi(0)\) \(\mathrm{x}=\mathrm{a}\) At \(\quad \mathrm{t}=3\), \(x=a \cos (3 \pi)\) \(\mathrm{x}=-\mathrm{a}\) Total displacement \((\Delta \mathrm{x})=\mathrm{a}-(-\mathrm{a})\) \(=2 \mathrm{a}\)
Kerala CEE - 2015
Motion in One Dimensions
141272
The time required to stop a car of mass \(800 \mathrm{~kg}\), moving at a speed of \(20 \mathrm{~ms}^{-1}\) over a distance of \(25 \mathrm{~m}\) is
1 \(2 \mathrm{~s}\)
2 \(2.5 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(4.5 \mathrm{~s}\)
5 \(1 \mathrm{~s}\)
Explanation:
B From the given question, we can acquire that, \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s}, \quad \mathrm{v}=0, \quad \mathrm{~s}=25 \mathrm{~m}\) From the equation, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \((0)^{2}-(20)^{2}=2 \times a \times 25\) \(-400=50 \times \mathrm{a}\) \(\mathrm{a}=-8 \mathrm{~m} / \mathrm{s}^{2}\) From the equation, \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(0=20-8 \mathrm{t}\) \(20=8 \mathrm{t}\) \(\mathrm{t}=\frac{20}{8}=2.5 \text { second }\)
Kerala CEE - 2015
Motion in One Dimensions
141273
A bullet when fired into a target loses half of its velocity after penetrating \(20 \mathrm{~cm}\). Further distance of penetration before it comes to rest is
1 \(6.66 \mathrm{~cm}\)
2 \(3.33 \mathrm{~cm}\)
3 \(12.5 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
5 \(5 \mathrm{~cm}\)
Explanation:
A Given that, initial velocity, \(\mathrm{u}=\mathrm{v}\) Final velocity \(\mathrm{v}=\mathrm{v} / 2\) Distance \((\mathrm{s})=20 \mathrm{~cm}\) Let the further distance of entry before coming to rest be \(\mathrm{x}\). \(v^{2}=u^{2}-2 a s\) \(\left(\frac{v}{2}\right)^{2}=v^{2}-2 a \times 20\) \(40 a=\frac{3 v^{2}}{4}\) \(a=\frac{3 v^{2}}{160}\) Again, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}^{2}=0+2 \mathrm{a} \times(20+\mathrm{x})\) Putting value of a from equation (i) \(v^{2}=2 \times \frac{3}{160} v^{2} \times(20+x)\) \(1=\frac{3}{80}(20+x)\) \(x=\frac{80-60}{3}\) \(x=6.66 \mathrm{~cm}\)
141267
A particle starts from rest and experiences constant acceleration for 6 seconds. If it travels a distance \(d_{1}\) in the first two seconds, a distance \(d_{2}\) in the next two seconds and a distance \(d_{3}\) in the last two seconds, then
C From the second equation of motion, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here \(\mathrm{u}=0\) So, \(\mathrm{s}=0+\frac{1}{2} \mathrm{at}^{2}=\frac{1}{2} \mathrm{at}^{2}\) Now according to the question, \(d_{1}=\frac{1}{2} a(2)^{2}=2 a\) \(d_{1}+d_{2}=\frac{1}{2} a(4)^{2}=8 a\) \(d_{2}=\left(d_{1}+d_{2}\right)-d_{1}=8 a-2 a=6 a\) \(d_{1}+d_{2}+d_{3}=\frac{1}{2} a(6)^{2}\) \(\because d_{3}=\left(d_{1}+d_{2}+d_{3}\right)-\left(d_{1}+d_{2}\right)\) \(d_{3}=\frac{1}{2} a(6)^{2}-8 a=10 a\) Hence, \(d_{1}: d_{2}: d_{3}=2 a: 6 a: 10 a\) \(=1: 3: 5\)
J and K-CET- 2003
Motion in One Dimensions
141269
A body is travelling east with a speed of \(9 \mathrm{~m} / \mathrm{s}\) and with an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) acting west on it. The displacement of the body during the \(5^{\text {th }}\) second of its motion is
1 \(0.25 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.75 \mathrm{~m}\)
4 zero
Explanation:
D Given, \(\mathrm{u}=9 \mathrm{~m} / \mathrm{s}\) east direction \(\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) west direction Distance covered in \(\mathrm{t}^{\text {th }}\) second (retardation)- \(\mathrm{s}_{\mathrm{t}}=\mathrm{u}-\frac{1}{2} \mathrm{a}(2 \mathrm{t}-1) \quad\left\{\because \mathrm{s}_{\mathrm{t}}=5 \mathrm{sec}\right\}\) \(\mathrm{s}_{5}=9-\frac{1}{2} \times 2(2 \times 5-1)=9-(10-1)\) \(\mathrm{s}_{5}=0\)
J and K-CET-2015
Motion in One Dimensions
141271
For a particle moving according to the equation \(x=a \cos \pi t\), the displacement in \(3 \mathrm{~s}\) is
1 0
2 \(0.5 \mathrm{a}\)
3 \(1.5 \mathrm{a}\)
4 \(2 \mathrm{a}\)
5 \(\mathrm{a}\)
Explanation:
D Given that, At \(\mathrm{t}=0, \mathrm{x}=\mathrm{a} \cos \pi(0)\) \(\mathrm{x}=\mathrm{a}\) At \(\quad \mathrm{t}=3\), \(x=a \cos (3 \pi)\) \(\mathrm{x}=-\mathrm{a}\) Total displacement \((\Delta \mathrm{x})=\mathrm{a}-(-\mathrm{a})\) \(=2 \mathrm{a}\)
Kerala CEE - 2015
Motion in One Dimensions
141272
The time required to stop a car of mass \(800 \mathrm{~kg}\), moving at a speed of \(20 \mathrm{~ms}^{-1}\) over a distance of \(25 \mathrm{~m}\) is
1 \(2 \mathrm{~s}\)
2 \(2.5 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(4.5 \mathrm{~s}\)
5 \(1 \mathrm{~s}\)
Explanation:
B From the given question, we can acquire that, \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s}, \quad \mathrm{v}=0, \quad \mathrm{~s}=25 \mathrm{~m}\) From the equation, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \((0)^{2}-(20)^{2}=2 \times a \times 25\) \(-400=50 \times \mathrm{a}\) \(\mathrm{a}=-8 \mathrm{~m} / \mathrm{s}^{2}\) From the equation, \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(0=20-8 \mathrm{t}\) \(20=8 \mathrm{t}\) \(\mathrm{t}=\frac{20}{8}=2.5 \text { second }\)
Kerala CEE - 2015
Motion in One Dimensions
141273
A bullet when fired into a target loses half of its velocity after penetrating \(20 \mathrm{~cm}\). Further distance of penetration before it comes to rest is
1 \(6.66 \mathrm{~cm}\)
2 \(3.33 \mathrm{~cm}\)
3 \(12.5 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
5 \(5 \mathrm{~cm}\)
Explanation:
A Given that, initial velocity, \(\mathrm{u}=\mathrm{v}\) Final velocity \(\mathrm{v}=\mathrm{v} / 2\) Distance \((\mathrm{s})=20 \mathrm{~cm}\) Let the further distance of entry before coming to rest be \(\mathrm{x}\). \(v^{2}=u^{2}-2 a s\) \(\left(\frac{v}{2}\right)^{2}=v^{2}-2 a \times 20\) \(40 a=\frac{3 v^{2}}{4}\) \(a=\frac{3 v^{2}}{160}\) Again, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}^{2}=0+2 \mathrm{a} \times(20+\mathrm{x})\) Putting value of a from equation (i) \(v^{2}=2 \times \frac{3}{160} v^{2} \times(20+x)\) \(1=\frac{3}{80}(20+x)\) \(x=\frac{80-60}{3}\) \(x=6.66 \mathrm{~cm}\)
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Motion in One Dimensions
141267
A particle starts from rest and experiences constant acceleration for 6 seconds. If it travels a distance \(d_{1}\) in the first two seconds, a distance \(d_{2}\) in the next two seconds and a distance \(d_{3}\) in the last two seconds, then
C From the second equation of motion, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here \(\mathrm{u}=0\) So, \(\mathrm{s}=0+\frac{1}{2} \mathrm{at}^{2}=\frac{1}{2} \mathrm{at}^{2}\) Now according to the question, \(d_{1}=\frac{1}{2} a(2)^{2}=2 a\) \(d_{1}+d_{2}=\frac{1}{2} a(4)^{2}=8 a\) \(d_{2}=\left(d_{1}+d_{2}\right)-d_{1}=8 a-2 a=6 a\) \(d_{1}+d_{2}+d_{3}=\frac{1}{2} a(6)^{2}\) \(\because d_{3}=\left(d_{1}+d_{2}+d_{3}\right)-\left(d_{1}+d_{2}\right)\) \(d_{3}=\frac{1}{2} a(6)^{2}-8 a=10 a\) Hence, \(d_{1}: d_{2}: d_{3}=2 a: 6 a: 10 a\) \(=1: 3: 5\)
J and K-CET- 2003
Motion in One Dimensions
141269
A body is travelling east with a speed of \(9 \mathrm{~m} / \mathrm{s}\) and with an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) acting west on it. The displacement of the body during the \(5^{\text {th }}\) second of its motion is
1 \(0.25 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.75 \mathrm{~m}\)
4 zero
Explanation:
D Given, \(\mathrm{u}=9 \mathrm{~m} / \mathrm{s}\) east direction \(\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) west direction Distance covered in \(\mathrm{t}^{\text {th }}\) second (retardation)- \(\mathrm{s}_{\mathrm{t}}=\mathrm{u}-\frac{1}{2} \mathrm{a}(2 \mathrm{t}-1) \quad\left\{\because \mathrm{s}_{\mathrm{t}}=5 \mathrm{sec}\right\}\) \(\mathrm{s}_{5}=9-\frac{1}{2} \times 2(2 \times 5-1)=9-(10-1)\) \(\mathrm{s}_{5}=0\)
J and K-CET-2015
Motion in One Dimensions
141271
For a particle moving according to the equation \(x=a \cos \pi t\), the displacement in \(3 \mathrm{~s}\) is
1 0
2 \(0.5 \mathrm{a}\)
3 \(1.5 \mathrm{a}\)
4 \(2 \mathrm{a}\)
5 \(\mathrm{a}\)
Explanation:
D Given that, At \(\mathrm{t}=0, \mathrm{x}=\mathrm{a} \cos \pi(0)\) \(\mathrm{x}=\mathrm{a}\) At \(\quad \mathrm{t}=3\), \(x=a \cos (3 \pi)\) \(\mathrm{x}=-\mathrm{a}\) Total displacement \((\Delta \mathrm{x})=\mathrm{a}-(-\mathrm{a})\) \(=2 \mathrm{a}\)
Kerala CEE - 2015
Motion in One Dimensions
141272
The time required to stop a car of mass \(800 \mathrm{~kg}\), moving at a speed of \(20 \mathrm{~ms}^{-1}\) over a distance of \(25 \mathrm{~m}\) is
1 \(2 \mathrm{~s}\)
2 \(2.5 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(4.5 \mathrm{~s}\)
5 \(1 \mathrm{~s}\)
Explanation:
B From the given question, we can acquire that, \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s}, \quad \mathrm{v}=0, \quad \mathrm{~s}=25 \mathrm{~m}\) From the equation, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \((0)^{2}-(20)^{2}=2 \times a \times 25\) \(-400=50 \times \mathrm{a}\) \(\mathrm{a}=-8 \mathrm{~m} / \mathrm{s}^{2}\) From the equation, \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(0=20-8 \mathrm{t}\) \(20=8 \mathrm{t}\) \(\mathrm{t}=\frac{20}{8}=2.5 \text { second }\)
Kerala CEE - 2015
Motion in One Dimensions
141273
A bullet when fired into a target loses half of its velocity after penetrating \(20 \mathrm{~cm}\). Further distance of penetration before it comes to rest is
1 \(6.66 \mathrm{~cm}\)
2 \(3.33 \mathrm{~cm}\)
3 \(12.5 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
5 \(5 \mathrm{~cm}\)
Explanation:
A Given that, initial velocity, \(\mathrm{u}=\mathrm{v}\) Final velocity \(\mathrm{v}=\mathrm{v} / 2\) Distance \((\mathrm{s})=20 \mathrm{~cm}\) Let the further distance of entry before coming to rest be \(\mathrm{x}\). \(v^{2}=u^{2}-2 a s\) \(\left(\frac{v}{2}\right)^{2}=v^{2}-2 a \times 20\) \(40 a=\frac{3 v^{2}}{4}\) \(a=\frac{3 v^{2}}{160}\) Again, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}^{2}=0+2 \mathrm{a} \times(20+\mathrm{x})\) Putting value of a from equation (i) \(v^{2}=2 \times \frac{3}{160} v^{2} \times(20+x)\) \(1=\frac{3}{80}(20+x)\) \(x=\frac{80-60}{3}\) \(x=6.66 \mathrm{~cm}\)
141267
A particle starts from rest and experiences constant acceleration for 6 seconds. If it travels a distance \(d_{1}\) in the first two seconds, a distance \(d_{2}\) in the next two seconds and a distance \(d_{3}\) in the last two seconds, then
C From the second equation of motion, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here \(\mathrm{u}=0\) So, \(\mathrm{s}=0+\frac{1}{2} \mathrm{at}^{2}=\frac{1}{2} \mathrm{at}^{2}\) Now according to the question, \(d_{1}=\frac{1}{2} a(2)^{2}=2 a\) \(d_{1}+d_{2}=\frac{1}{2} a(4)^{2}=8 a\) \(d_{2}=\left(d_{1}+d_{2}\right)-d_{1}=8 a-2 a=6 a\) \(d_{1}+d_{2}+d_{3}=\frac{1}{2} a(6)^{2}\) \(\because d_{3}=\left(d_{1}+d_{2}+d_{3}\right)-\left(d_{1}+d_{2}\right)\) \(d_{3}=\frac{1}{2} a(6)^{2}-8 a=10 a\) Hence, \(d_{1}: d_{2}: d_{3}=2 a: 6 a: 10 a\) \(=1: 3: 5\)
J and K-CET- 2003
Motion in One Dimensions
141269
A body is travelling east with a speed of \(9 \mathrm{~m} / \mathrm{s}\) and with an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) acting west on it. The displacement of the body during the \(5^{\text {th }}\) second of its motion is
1 \(0.25 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.75 \mathrm{~m}\)
4 zero
Explanation:
D Given, \(\mathrm{u}=9 \mathrm{~m} / \mathrm{s}\) east direction \(\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) west direction Distance covered in \(\mathrm{t}^{\text {th }}\) second (retardation)- \(\mathrm{s}_{\mathrm{t}}=\mathrm{u}-\frac{1}{2} \mathrm{a}(2 \mathrm{t}-1) \quad\left\{\because \mathrm{s}_{\mathrm{t}}=5 \mathrm{sec}\right\}\) \(\mathrm{s}_{5}=9-\frac{1}{2} \times 2(2 \times 5-1)=9-(10-1)\) \(\mathrm{s}_{5}=0\)
J and K-CET-2015
Motion in One Dimensions
141271
For a particle moving according to the equation \(x=a \cos \pi t\), the displacement in \(3 \mathrm{~s}\) is
1 0
2 \(0.5 \mathrm{a}\)
3 \(1.5 \mathrm{a}\)
4 \(2 \mathrm{a}\)
5 \(\mathrm{a}\)
Explanation:
D Given that, At \(\mathrm{t}=0, \mathrm{x}=\mathrm{a} \cos \pi(0)\) \(\mathrm{x}=\mathrm{a}\) At \(\quad \mathrm{t}=3\), \(x=a \cos (3 \pi)\) \(\mathrm{x}=-\mathrm{a}\) Total displacement \((\Delta \mathrm{x})=\mathrm{a}-(-\mathrm{a})\) \(=2 \mathrm{a}\)
Kerala CEE - 2015
Motion in One Dimensions
141272
The time required to stop a car of mass \(800 \mathrm{~kg}\), moving at a speed of \(20 \mathrm{~ms}^{-1}\) over a distance of \(25 \mathrm{~m}\) is
1 \(2 \mathrm{~s}\)
2 \(2.5 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(4.5 \mathrm{~s}\)
5 \(1 \mathrm{~s}\)
Explanation:
B From the given question, we can acquire that, \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s}, \quad \mathrm{v}=0, \quad \mathrm{~s}=25 \mathrm{~m}\) From the equation, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \((0)^{2}-(20)^{2}=2 \times a \times 25\) \(-400=50 \times \mathrm{a}\) \(\mathrm{a}=-8 \mathrm{~m} / \mathrm{s}^{2}\) From the equation, \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(0=20-8 \mathrm{t}\) \(20=8 \mathrm{t}\) \(\mathrm{t}=\frac{20}{8}=2.5 \text { second }\)
Kerala CEE - 2015
Motion in One Dimensions
141273
A bullet when fired into a target loses half of its velocity after penetrating \(20 \mathrm{~cm}\). Further distance of penetration before it comes to rest is
1 \(6.66 \mathrm{~cm}\)
2 \(3.33 \mathrm{~cm}\)
3 \(12.5 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
5 \(5 \mathrm{~cm}\)
Explanation:
A Given that, initial velocity, \(\mathrm{u}=\mathrm{v}\) Final velocity \(\mathrm{v}=\mathrm{v} / 2\) Distance \((\mathrm{s})=20 \mathrm{~cm}\) Let the further distance of entry before coming to rest be \(\mathrm{x}\). \(v^{2}=u^{2}-2 a s\) \(\left(\frac{v}{2}\right)^{2}=v^{2}-2 a \times 20\) \(40 a=\frac{3 v^{2}}{4}\) \(a=\frac{3 v^{2}}{160}\) Again, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}^{2}=0+2 \mathrm{a} \times(20+\mathrm{x})\) Putting value of a from equation (i) \(v^{2}=2 \times \frac{3}{160} v^{2} \times(20+x)\) \(1=\frac{3}{80}(20+x)\) \(x=\frac{80-60}{3}\) \(x=6.66 \mathrm{~cm}\)
141267
A particle starts from rest and experiences constant acceleration for 6 seconds. If it travels a distance \(d_{1}\) in the first two seconds, a distance \(d_{2}\) in the next two seconds and a distance \(d_{3}\) in the last two seconds, then
C From the second equation of motion, \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here \(\mathrm{u}=0\) So, \(\mathrm{s}=0+\frac{1}{2} \mathrm{at}^{2}=\frac{1}{2} \mathrm{at}^{2}\) Now according to the question, \(d_{1}=\frac{1}{2} a(2)^{2}=2 a\) \(d_{1}+d_{2}=\frac{1}{2} a(4)^{2}=8 a\) \(d_{2}=\left(d_{1}+d_{2}\right)-d_{1}=8 a-2 a=6 a\) \(d_{1}+d_{2}+d_{3}=\frac{1}{2} a(6)^{2}\) \(\because d_{3}=\left(d_{1}+d_{2}+d_{3}\right)-\left(d_{1}+d_{2}\right)\) \(d_{3}=\frac{1}{2} a(6)^{2}-8 a=10 a\) Hence, \(d_{1}: d_{2}: d_{3}=2 a: 6 a: 10 a\) \(=1: 3: 5\)
J and K-CET- 2003
Motion in One Dimensions
141269
A body is travelling east with a speed of \(9 \mathrm{~m} / \mathrm{s}\) and with an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) acting west on it. The displacement of the body during the \(5^{\text {th }}\) second of its motion is
1 \(0.25 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.75 \mathrm{~m}\)
4 zero
Explanation:
D Given, \(\mathrm{u}=9 \mathrm{~m} / \mathrm{s}\) east direction \(\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) west direction Distance covered in \(\mathrm{t}^{\text {th }}\) second (retardation)- \(\mathrm{s}_{\mathrm{t}}=\mathrm{u}-\frac{1}{2} \mathrm{a}(2 \mathrm{t}-1) \quad\left\{\because \mathrm{s}_{\mathrm{t}}=5 \mathrm{sec}\right\}\) \(\mathrm{s}_{5}=9-\frac{1}{2} \times 2(2 \times 5-1)=9-(10-1)\) \(\mathrm{s}_{5}=0\)
J and K-CET-2015
Motion in One Dimensions
141271
For a particle moving according to the equation \(x=a \cos \pi t\), the displacement in \(3 \mathrm{~s}\) is
1 0
2 \(0.5 \mathrm{a}\)
3 \(1.5 \mathrm{a}\)
4 \(2 \mathrm{a}\)
5 \(\mathrm{a}\)
Explanation:
D Given that, At \(\mathrm{t}=0, \mathrm{x}=\mathrm{a} \cos \pi(0)\) \(\mathrm{x}=\mathrm{a}\) At \(\quad \mathrm{t}=3\), \(x=a \cos (3 \pi)\) \(\mathrm{x}=-\mathrm{a}\) Total displacement \((\Delta \mathrm{x})=\mathrm{a}-(-\mathrm{a})\) \(=2 \mathrm{a}\)
Kerala CEE - 2015
Motion in One Dimensions
141272
The time required to stop a car of mass \(800 \mathrm{~kg}\), moving at a speed of \(20 \mathrm{~ms}^{-1}\) over a distance of \(25 \mathrm{~m}\) is
1 \(2 \mathrm{~s}\)
2 \(2.5 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(4.5 \mathrm{~s}\)
5 \(1 \mathrm{~s}\)
Explanation:
B From the given question, we can acquire that, \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s}, \quad \mathrm{v}=0, \quad \mathrm{~s}=25 \mathrm{~m}\) From the equation, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \((0)^{2}-(20)^{2}=2 \times a \times 25\) \(-400=50 \times \mathrm{a}\) \(\mathrm{a}=-8 \mathrm{~m} / \mathrm{s}^{2}\) From the equation, \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(0=20-8 \mathrm{t}\) \(20=8 \mathrm{t}\) \(\mathrm{t}=\frac{20}{8}=2.5 \text { second }\)
Kerala CEE - 2015
Motion in One Dimensions
141273
A bullet when fired into a target loses half of its velocity after penetrating \(20 \mathrm{~cm}\). Further distance of penetration before it comes to rest is
1 \(6.66 \mathrm{~cm}\)
2 \(3.33 \mathrm{~cm}\)
3 \(12.5 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
5 \(5 \mathrm{~cm}\)
Explanation:
A Given that, initial velocity, \(\mathrm{u}=\mathrm{v}\) Final velocity \(\mathrm{v}=\mathrm{v} / 2\) Distance \((\mathrm{s})=20 \mathrm{~cm}\) Let the further distance of entry before coming to rest be \(\mathrm{x}\). \(v^{2}=u^{2}-2 a s\) \(\left(\frac{v}{2}\right)^{2}=v^{2}-2 a \times 20\) \(40 a=\frac{3 v^{2}}{4}\) \(a=\frac{3 v^{2}}{160}\) Again, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}^{2}=0+2 \mathrm{a} \times(20+\mathrm{x})\) Putting value of a from equation (i) \(v^{2}=2 \times \frac{3}{160} v^{2} \times(20+x)\) \(1=\frac{3}{80}(20+x)\) \(x=\frac{80-60}{3}\) \(x=6.66 \mathrm{~cm}\)