362328
A bullet fired into a fixed target loses half of its velocity after penetrating \(3\,cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
1 \(1.0\;cm\)
2 \(1.5\;cm\)
3 \(2.0\;cm\)
4 \(3.0\;cm\)
Explanation:
Let initial velocity of the bullet \( = u\) After penetrating \(3\,cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a(3)\) \( \Rightarrow 6a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{{u^2}}}{8}\) Further it will penetrate through distance \(x\) For distance \(x,\;v = 0,\;u = u/2,\;s = x,\;a = {u^2}/8\) From \({v^2} = {u^2} - 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2{\left( {\frac{u}{8}} \right)^2} \cdot x\) \( \Rightarrow x = 1\;cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362329
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the \({5^{th}}\) sec to that covered in 5 sec is
1 \(1/25\)
2 \(25/9\)
3 \(9/25\)
4 \(3/5\)
Explanation:
Distance covered by the body \({5^{th}}\) second. \({S_{{5^{th}}}} = ut + \frac{a}{2}\left( {2n - 1} \right) = 0 + \frac{a}{2}\left( {2 \times 5 - 1} \right) = \frac{{9a}}{2}\) and distance covered in 5 second, \({S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2}a \times 25 = \frac{{25a}}{2}\) \(\therefore \;\frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362330
A bullet fired into a fixed target loses half of its velocity after penetrating 1 \(cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
1 \(2.0\,cm\)
2 \(1.5\,cm\)
3 \(\frac{2}{3}\,cm\)
4 \(\frac{1}{3}\,cm\)
Explanation:
Let initial velocity of the bullet be \(u\) After penetrating 1 \(cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a\left( 1 \right)\) \( \Rightarrow 2a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{3{u^2}}}{8}\) Further it will penetrate through distance \(x\) and stops. For distance \(x,v = 0,\,u = u/2,s = x,a = 3{u^2}/8\) From \({v^2} - {u^2} = 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2\left( {\frac{{3{u^2}}}{8}} \right) \cdot x\) \( \Rightarrow x = \frac{1}{3} \, cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362331
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \,ms^{-2}\), in the third second is
1 \(6\,m\)
2 \(4\,m\)
3 \(\frac{{10}}{3}\,m\)
4 \(\frac{{19}}{3}\,m\)
Explanation:
Distance travelled in the nth second is given by \(d = u + \frac{a}{2}\left(2n - 1 \right)\) Put \(u = 0, \; a = \frac{4}{3} \, ms^{-2}, \; n = 3\) \(\therefore d = 0 + \frac{4}{{3 \times 2}}\left( {2 \times 3 - 1} \right) = \frac{4}{6} \times 5 = \frac{{10}}{3} \, m\)
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PHXI03:MOTION IN A STRAIGHT LINE
362328
A bullet fired into a fixed target loses half of its velocity after penetrating \(3\,cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
1 \(1.0\;cm\)
2 \(1.5\;cm\)
3 \(2.0\;cm\)
4 \(3.0\;cm\)
Explanation:
Let initial velocity of the bullet \( = u\) After penetrating \(3\,cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a(3)\) \( \Rightarrow 6a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{{u^2}}}{8}\) Further it will penetrate through distance \(x\) For distance \(x,\;v = 0,\;u = u/2,\;s = x,\;a = {u^2}/8\) From \({v^2} = {u^2} - 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2{\left( {\frac{u}{8}} \right)^2} \cdot x\) \( \Rightarrow x = 1\;cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362329
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the \({5^{th}}\) sec to that covered in 5 sec is
1 \(1/25\)
2 \(25/9\)
3 \(9/25\)
4 \(3/5\)
Explanation:
Distance covered by the body \({5^{th}}\) second. \({S_{{5^{th}}}} = ut + \frac{a}{2}\left( {2n - 1} \right) = 0 + \frac{a}{2}\left( {2 \times 5 - 1} \right) = \frac{{9a}}{2}\) and distance covered in 5 second, \({S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2}a \times 25 = \frac{{25a}}{2}\) \(\therefore \;\frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362330
A bullet fired into a fixed target loses half of its velocity after penetrating 1 \(cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
1 \(2.0\,cm\)
2 \(1.5\,cm\)
3 \(\frac{2}{3}\,cm\)
4 \(\frac{1}{3}\,cm\)
Explanation:
Let initial velocity of the bullet be \(u\) After penetrating 1 \(cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a\left( 1 \right)\) \( \Rightarrow 2a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{3{u^2}}}{8}\) Further it will penetrate through distance \(x\) and stops. For distance \(x,v = 0,\,u = u/2,s = x,a = 3{u^2}/8\) From \({v^2} - {u^2} = 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2\left( {\frac{{3{u^2}}}{8}} \right) \cdot x\) \( \Rightarrow x = \frac{1}{3} \, cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362331
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \,ms^{-2}\), in the third second is
1 \(6\,m\)
2 \(4\,m\)
3 \(\frac{{10}}{3}\,m\)
4 \(\frac{{19}}{3}\,m\)
Explanation:
Distance travelled in the nth second is given by \(d = u + \frac{a}{2}\left(2n - 1 \right)\) Put \(u = 0, \; a = \frac{4}{3} \, ms^{-2}, \; n = 3\) \(\therefore d = 0 + \frac{4}{{3 \times 2}}\left( {2 \times 3 - 1} \right) = \frac{4}{6} \times 5 = \frac{{10}}{3} \, m\)
362328
A bullet fired into a fixed target loses half of its velocity after penetrating \(3\,cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
1 \(1.0\;cm\)
2 \(1.5\;cm\)
3 \(2.0\;cm\)
4 \(3.0\;cm\)
Explanation:
Let initial velocity of the bullet \( = u\) After penetrating \(3\,cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a(3)\) \( \Rightarrow 6a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{{u^2}}}{8}\) Further it will penetrate through distance \(x\) For distance \(x,\;v = 0,\;u = u/2,\;s = x,\;a = {u^2}/8\) From \({v^2} = {u^2} - 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2{\left( {\frac{u}{8}} \right)^2} \cdot x\) \( \Rightarrow x = 1\;cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362329
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the \({5^{th}}\) sec to that covered in 5 sec is
1 \(1/25\)
2 \(25/9\)
3 \(9/25\)
4 \(3/5\)
Explanation:
Distance covered by the body \({5^{th}}\) second. \({S_{{5^{th}}}} = ut + \frac{a}{2}\left( {2n - 1} \right) = 0 + \frac{a}{2}\left( {2 \times 5 - 1} \right) = \frac{{9a}}{2}\) and distance covered in 5 second, \({S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2}a \times 25 = \frac{{25a}}{2}\) \(\therefore \;\frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362330
A bullet fired into a fixed target loses half of its velocity after penetrating 1 \(cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
1 \(2.0\,cm\)
2 \(1.5\,cm\)
3 \(\frac{2}{3}\,cm\)
4 \(\frac{1}{3}\,cm\)
Explanation:
Let initial velocity of the bullet be \(u\) After penetrating 1 \(cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a\left( 1 \right)\) \( \Rightarrow 2a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{3{u^2}}}{8}\) Further it will penetrate through distance \(x\) and stops. For distance \(x,v = 0,\,u = u/2,s = x,a = 3{u^2}/8\) From \({v^2} - {u^2} = 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2\left( {\frac{{3{u^2}}}{8}} \right) \cdot x\) \( \Rightarrow x = \frac{1}{3} \, cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362331
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \,ms^{-2}\), in the third second is
1 \(6\,m\)
2 \(4\,m\)
3 \(\frac{{10}}{3}\,m\)
4 \(\frac{{19}}{3}\,m\)
Explanation:
Distance travelled in the nth second is given by \(d = u + \frac{a}{2}\left(2n - 1 \right)\) Put \(u = 0, \; a = \frac{4}{3} \, ms^{-2}, \; n = 3\) \(\therefore d = 0 + \frac{4}{{3 \times 2}}\left( {2 \times 3 - 1} \right) = \frac{4}{6} \times 5 = \frac{{10}}{3} \, m\)
362328
A bullet fired into a fixed target loses half of its velocity after penetrating \(3\,cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
1 \(1.0\;cm\)
2 \(1.5\;cm\)
3 \(2.0\;cm\)
4 \(3.0\;cm\)
Explanation:
Let initial velocity of the bullet \( = u\) After penetrating \(3\,cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a(3)\) \( \Rightarrow 6a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{{u^2}}}{8}\) Further it will penetrate through distance \(x\) For distance \(x,\;v = 0,\;u = u/2,\;s = x,\;a = {u^2}/8\) From \({v^2} = {u^2} - 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2{\left( {\frac{u}{8}} \right)^2} \cdot x\) \( \Rightarrow x = 1\;cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362329
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the \({5^{th}}\) sec to that covered in 5 sec is
1 \(1/25\)
2 \(25/9\)
3 \(9/25\)
4 \(3/5\)
Explanation:
Distance covered by the body \({5^{th}}\) second. \({S_{{5^{th}}}} = ut + \frac{a}{2}\left( {2n - 1} \right) = 0 + \frac{a}{2}\left( {2 \times 5 - 1} \right) = \frac{{9a}}{2}\) and distance covered in 5 second, \({S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2}a \times 25 = \frac{{25a}}{2}\) \(\therefore \;\frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362330
A bullet fired into a fixed target loses half of its velocity after penetrating 1 \(cm\). How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
1 \(2.0\,cm\)
2 \(1.5\,cm\)
3 \(\frac{2}{3}\,cm\)
4 \(\frac{1}{3}\,cm\)
Explanation:
Let initial velocity of the bullet be \(u\) After penetrating 1 \(cm\) its velocity becomes \(\frac{u}{2}\) From \({v^2} = {u^2} - 2as\) \({\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a\left( 1 \right)\) \( \Rightarrow 2a = \frac{{3{u^2}}}{4} \Rightarrow a = \frac{{3{u^2}}}{8}\) Further it will penetrate through distance \(x\) and stops. For distance \(x,v = 0,\,u = u/2,s = x,a = 3{u^2}/8\) From \({v^2} - {u^2} = 2as \Rightarrow 0 = {\left( {\frac{u}{2}} \right)^2} - 2\left( {\frac{{3{u^2}}}{8}} \right) \cdot x\) \( \Rightarrow x = \frac{1}{3} \, cm\)
PHXI03:MOTION IN A STRAIGHT LINE
362331
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \,ms^{-2}\), in the third second is
1 \(6\,m\)
2 \(4\,m\)
3 \(\frac{{10}}{3}\,m\)
4 \(\frac{{19}}{3}\,m\)
Explanation:
Distance travelled in the nth second is given by \(d = u + \frac{a}{2}\left(2n - 1 \right)\) Put \(u = 0, \; a = \frac{4}{3} \, ms^{-2}, \; n = 3\) \(\therefore d = 0 + \frac{4}{{3 \times 2}}\left( {2 \times 3 - 1} \right) = \frac{4}{6} \times 5 = \frac{{10}}{3} \, m\)