03. Equation of Motion
Motion in One Dimensions

141754 A body of mass \(10 \mathrm{~kg}\) is moving with a constant velocity of \(10 \mathrm{~m} / \mathrm{s}\). When a constant force acts for \(4 \mathrm{~s}\) on it, it moves with a velocity \(2 \mathrm{~m} / \mathrm{s}\) in the opposite direction. The acceleration produced in it is

1 \(3 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(-3 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(0.3 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(0.03 \mathrm{~m} / \mathrm{s}^{2}\)
Motion in One Dimensions

141755 A body starts from rest with uniform acceleration. If its velocity after \(n\) second is \(v\), then its displacement in the last \(2 \mathrm{~s}\) is

1 \(\frac{2 v(n+1)}{n}\)
2 \(\frac{\mathrm{v}(\mathrm{n}+1)}{\mathrm{n}}\)
3 \(\frac{\mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
4 \(\frac{2 \mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
Motion in One Dimensions

141756 An automobile travelling at \(50 \mathrm{~km} / \mathrm{h}\), can be stopped at a distance of \(40 \mathrm{~m}\) by applying brakes. If the same automobile is travelling at \(90 \mathrm{~km} / \mathrm{h}\), all other conditions remaining same and assuming no skidding, the minimum stopping distance in metres is :

1 72
2 92.5
3 102.6
4 129.6
Motion in One Dimensions

141757 The velocity of a body moving with uniform acceleration at a given instant of time \(t\) is \(\mathbf{1 0}\) \(\mathrm{m} / \mathrm{s}\). After \(5 \mathrm{~s}\) its velocity is \(20 \mathrm{~m} / \mathrm{s}\). Distance travelled in that time is:

1 \(300 \mathrm{~m}\)
2 \(400 \mathrm{~m}\)
3 \(150 \mathrm{~m}\)
4 \(75 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Motion in One Dimensions

141754 A body of mass \(10 \mathrm{~kg}\) is moving with a constant velocity of \(10 \mathrm{~m} / \mathrm{s}\). When a constant force acts for \(4 \mathrm{~s}\) on it, it moves with a velocity \(2 \mathrm{~m} / \mathrm{s}\) in the opposite direction. The acceleration produced in it is

1 \(3 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(-3 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(0.3 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(0.03 \mathrm{~m} / \mathrm{s}^{2}\)
Motion in One Dimensions

141755 A body starts from rest with uniform acceleration. If its velocity after \(n\) second is \(v\), then its displacement in the last \(2 \mathrm{~s}\) is

1 \(\frac{2 v(n+1)}{n}\)
2 \(\frac{\mathrm{v}(\mathrm{n}+1)}{\mathrm{n}}\)
3 \(\frac{\mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
4 \(\frac{2 \mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
Motion in One Dimensions

141756 An automobile travelling at \(50 \mathrm{~km} / \mathrm{h}\), can be stopped at a distance of \(40 \mathrm{~m}\) by applying brakes. If the same automobile is travelling at \(90 \mathrm{~km} / \mathrm{h}\), all other conditions remaining same and assuming no skidding, the minimum stopping distance in metres is :

1 72
2 92.5
3 102.6
4 129.6
Motion in One Dimensions

141757 The velocity of a body moving with uniform acceleration at a given instant of time \(t\) is \(\mathbf{1 0}\) \(\mathrm{m} / \mathrm{s}\). After \(5 \mathrm{~s}\) its velocity is \(20 \mathrm{~m} / \mathrm{s}\). Distance travelled in that time is:

1 \(300 \mathrm{~m}\)
2 \(400 \mathrm{~m}\)
3 \(150 \mathrm{~m}\)
4 \(75 \mathrm{~m}\)
Motion in One Dimensions

141754 A body of mass \(10 \mathrm{~kg}\) is moving with a constant velocity of \(10 \mathrm{~m} / \mathrm{s}\). When a constant force acts for \(4 \mathrm{~s}\) on it, it moves with a velocity \(2 \mathrm{~m} / \mathrm{s}\) in the opposite direction. The acceleration produced in it is

1 \(3 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(-3 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(0.3 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(0.03 \mathrm{~m} / \mathrm{s}^{2}\)
Motion in One Dimensions

141755 A body starts from rest with uniform acceleration. If its velocity after \(n\) second is \(v\), then its displacement in the last \(2 \mathrm{~s}\) is

1 \(\frac{2 v(n+1)}{n}\)
2 \(\frac{\mathrm{v}(\mathrm{n}+1)}{\mathrm{n}}\)
3 \(\frac{\mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
4 \(\frac{2 \mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
Motion in One Dimensions

141756 An automobile travelling at \(50 \mathrm{~km} / \mathrm{h}\), can be stopped at a distance of \(40 \mathrm{~m}\) by applying brakes. If the same automobile is travelling at \(90 \mathrm{~km} / \mathrm{h}\), all other conditions remaining same and assuming no skidding, the minimum stopping distance in metres is :

1 72
2 92.5
3 102.6
4 129.6
Motion in One Dimensions

141757 The velocity of a body moving with uniform acceleration at a given instant of time \(t\) is \(\mathbf{1 0}\) \(\mathrm{m} / \mathrm{s}\). After \(5 \mathrm{~s}\) its velocity is \(20 \mathrm{~m} / \mathrm{s}\). Distance travelled in that time is:

1 \(300 \mathrm{~m}\)
2 \(400 \mathrm{~m}\)
3 \(150 \mathrm{~m}\)
4 \(75 \mathrm{~m}\)
Motion in One Dimensions

141754 A body of mass \(10 \mathrm{~kg}\) is moving with a constant velocity of \(10 \mathrm{~m} / \mathrm{s}\). When a constant force acts for \(4 \mathrm{~s}\) on it, it moves with a velocity \(2 \mathrm{~m} / \mathrm{s}\) in the opposite direction. The acceleration produced in it is

1 \(3 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(-3 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(0.3 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(0.03 \mathrm{~m} / \mathrm{s}^{2}\)
Motion in One Dimensions

141755 A body starts from rest with uniform acceleration. If its velocity after \(n\) second is \(v\), then its displacement in the last \(2 \mathrm{~s}\) is

1 \(\frac{2 v(n+1)}{n}\)
2 \(\frac{\mathrm{v}(\mathrm{n}+1)}{\mathrm{n}}\)
3 \(\frac{\mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
4 \(\frac{2 \mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}\)
Motion in One Dimensions

141756 An automobile travelling at \(50 \mathrm{~km} / \mathrm{h}\), can be stopped at a distance of \(40 \mathrm{~m}\) by applying brakes. If the same automobile is travelling at \(90 \mathrm{~km} / \mathrm{h}\), all other conditions remaining same and assuming no skidding, the minimum stopping distance in metres is :

1 72
2 92.5
3 102.6
4 129.6
Motion in One Dimensions

141757 The velocity of a body moving with uniform acceleration at a given instant of time \(t\) is \(\mathbf{1 0}\) \(\mathrm{m} / \mathrm{s}\). After \(5 \mathrm{~s}\) its velocity is \(20 \mathrm{~m} / \mathrm{s}\). Distance travelled in that time is:

1 \(300 \mathrm{~m}\)
2 \(400 \mathrm{~m}\)
3 \(150 \mathrm{~m}\)
4 \(75 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here