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>>Motion in a Straight Line
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Motion In A Straight Line

Q: Two cars are travelling towards each other on a straight road at velocities \$15\ m/s\$ and \$16\ m/s\$ respectively. When they are \$150\ m\$ apart, both the drivers apply the brakes and the cars decelerate at \$3\ m/s^{2}\$ and \$4\ m/s^{2}\$ until they stop. Separation between the cars when they come to rest is :

We know, \$v^2=u^2-2as\$Given, Initial velocity of car 1, \$u_1=15\ m/s\$, Acceleration, \$a=3\ m/s^2\$ Initial velocity of car 2, \$u_2=16\ m/s\$, Acceleration, \$a=4\ m/s^2\$For car 1,Final Velocity, \$v_1 = 0\ m/s\$\$-15^2 = -2(3) \times s\$\$\therefore s_1= 37.5\$ mFor car 2,Final Velocity, \$v_2 = 0\ m/s\$\$-16^2 = -2(4) \times s\$\$\therefore s_2= 32\$ mTotal displacement,\$s_1+s_2 = 37.5+32 = 69.5\$ m\$\therefore\$ separation between them,\$= 150-69.5 =80.5\$ m

Q: The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:

We have average velocity \$= \dfrac{g}{2}\$ Since it is a free fall, average velocity, \$v=\dfrac{S}{t}= \dfrac{\dfrac{1}{2}gt^2}{t} = \dfrac{g}{2}\$, where \$t\$ is the time of descent and initial velocity is zero.We know that for a free fall, velocity with which it reaches the ground: \$V = gt\$, where \$t\$ is time of descent . Substituting the same in the above equation for average velocity: \$\dfrac{V}{2} = \dfrac{g}{2} \$\$\therefore V=g\$

Q: A body leaving a certain point \$O\$ moves with an acceleration which is constant. At the end of the first second after it left \$O\$, its velocity is \$1.5m/\sec\$. At the end of the sixth second after it left \$O\$ the body stops momentarily and begins to move backwards. The distance traveled by the body before it stops momentarily is

\$v=u+at\$\$1.5=u+5a...........(i)\$\$0=u+6a.............(ii)\$From (i) and (ii)\$0=1.5+a(6-5)\$\$a=-1.5m/s^2\$Om substituting \$a=-1.5m/s^2\$ in equation (ii)\$0=u+6(-1.5)\$\$0=u-9\$\$u=9m/s\$The distance traveled by the body in its forward journey is \$s_1\$ can be calculated as follows,\$s_1=ut+\frac{1}{2}at^2\$\$s_1=9\times6+\frac{1}{2}\times(-1.5)\times6^2\$\$s_1=54+(-1.5\times18)\$\$s_1=27m\$The distance traveled by the body in its forward journey \$(s_1)\$= the distance traveled by the body in its backward journey\$(s_2)=27m\$So, the total distance traveled by the body in forward journey and its backward journey\$=s_1+s_2=27+27=54m\$

Q: A ball dropped from one metre above the top of a window, crosses the window in \$t_{1}\;s\$ . If the same ball is dropped from \$2\;m\$ above the top of the same window, time taken by it to cross the window is \$t_{2}\; s\$ . Then

Solution 1The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:\$v^2=u^2+2aS\$, where \$u=0\$The ball with higher speed will take less time to cross the window.\$\because\$ \$H_2 > H_1\$ \$\implies\$ \$v_2 > v_1\$ \$\therefore\$ \$t_2 < t_1\$Hence Option D is correctSolution 2Statement D is correctSituation 1:We want time to cross x.\$v^{2}=u^{2}+2as\$\$u=0, a=-g, s=-1 \$\$v^{2}=2g\Rightarrow v_{1}=\sqrt{2g}\$\$(1+x)=\dfrac{1}{2} gt^{2}\RightarrowT_{2}=\sqrt{\dfrac{2(1+x)}{g}}\$\$1=\dfrac{1}{2}gt^{2}=T_{1}=\sqrt{\dfrac{2}{g}}\$So time taken to cross \$x=t_{1}=T_{2}-T_{1}=\sqrt{\dfrac{2}{g}}(\sqrt{1+x}-1)\$Situation 2:\$(2+x)=\dfrac{1}{2}gt^{2}\RightarrowT_{2}=\sqrt{\dfrac{2(2+x)}{g}}\$\$2=\dfrac{1}{2}gt^{2}\Rightarrow T_{1}=\sqrt{\dfrac{2\times 2}{g}}\$So time taken to cross \$x=t_{2}=T_{2}-T_{1}=\sqrt{\dfrac{2}{g}}(\sqrt{2+x}-\sqrt{2})\$Clearly \$t_{2}< t_{1} \$

Q: A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity \$4\times 10^3\$ \$km/s\$ at a distance of \$2\;cm\$, the time required to reach the given velocity is :

\$v^2 = u^2 + 2as\$\$u = 0, s = 2 \times 10^{-2} mt\$\$a = \dfrac {v^2}{2s} = \dfrac {(4 \times 10^3 \times 10^3)^2}{2 \times 2 \times 10^{-2}} m/s^2\$\$a = 4 \times 10^{14} \$Now, \$ v = u + at\$\$\dfrac {v}{a} = t\$\$t = \dfrac{4 \times 10^6}{4 \times 10^{14}} = 10^{-8}s\$

Q: From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If \$V_{A}\$ and \$V _{B}\$ are their respective velocities on reaching the ground then :

For A:\$s_A = vt_1 - \dfrac {1}{2} gt_1^2\$\$s_B = -vt_2 - \dfrac {1}{2} gt_2^2\$We have, \$s_A = s_B\$So, on solving we get:\$t_2 - t_1 = \dfrac {2v}{g}\$ ...........(1)Also,\$v_A = u_A - gt_1\$\$v_B = u_B - gt_2\$Using \$(1)\$, we get:\$ \dfrac {v_A}{v_B} = \dfrac {v - gt_1}{-v -gt_2}\$\$= \dfrac {v - gt_1}{-v -g (t_1 + \dfrac {2v}{g})} = 1\$\$\Rightarrow v_A = v_B\$

Q: It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minutes to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator?

Let the distance to the top be d. therefore the speed of the man with escalator not moving \$=(\dfrac{d}{1}) m/min\$ therefore the speed of the escalator with the man at rest\$=(\dfrac{d}{3})\$m/min when both the main and the escalator are moving, effective speed\$= (\dfrac{d}{3} +d)\$ m/min\$= \dfrac{4d}{3}\$ m/min Therefore time taken when both men and escalator are moving\$= \dfrac{distance}{speed} = \dfrac{d}{\dfrac{4d}{3}}\$\$\dfrac{3}{4} min = 45\$secondsHence (B) is correct answer

Q: The average velocity of a body moving with uniform acceleration after travelling a distance of \$3.06\ m\$ is \$0.34\ {m/s}\$. The change in velocity of the body is \$0.18\ {m/s}\$. During this time, its acceleration is

\$V_{avg} = \dfrac {D}{t} \Rightarrow 0.34 = \dfrac {3.06}{t}\$\$\Rightarrow t = \dfrac {3.06}{0.34}\$ = \$9\ sec\$We have, \$v = u + at\$\$v - u = at\$\$0.18 = a \times 9\$\$\Rightarrow a = \dfrac {0.18}{9} = 0.02 \ m/s^2\$

Q: A body travels \$200\;m\$ in the first two second and \$220\;m\$ in the next four second. The velocity at the end of the seventh second from the start will be

\$200 = u(2) + \dfrac {1}{2} \times a (2)^2 = 2u + 2a = 200\$\$\Rightarrow u + a =100 - (1)\$Distance in next \$4\ sec\$ \$ \Rightarrow s_6 - s_2\$\$\Rightarrow 220 = (u (6)] + \dfrac {1}{2} a (6)^2) - (u (2) + \dfrac {1}{2} a (2)^2)\$\$420 = 6u + 18a - (2)\$\$420 = 6 (100 - a) + 18a (from 1)\$\$\Rightarrow -180 = 12a \Rightarrow a = - 15 m/s^2 , u = 115 m/s\$\$v\ (at \;end \;of \;7 \;sec) = u + at\$\$= 115 + (-15 \times 7) = 10 m/s\$

Physics
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Easy Qs Med Qs Hard Qs

Two cars are travelling towards each other on a straight road at velocities $15 m / s$ and $16 m / s$ respectively. When they are $150 m$ apart, both the drivers apply the brakes and the cars decelerate at $3 m / s^{2}$ and $4 m / s^{2}$ until they stop. Separation between the cars when they come to rest is :

86.5 m

89.5 m

85.5 m

80.5 m

Medium

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The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:

g

\frac{g}{2}

\frac{g}{2}

2 g

Medium

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A body leaving a certain point $O$ moves with an acceleration which is constant. At the end of the first second after it left $O$ , its velocity is $1.5 m / sec$ . At the end of the sixth second after it left $O$ the body stops momentarily and begins to move backwards. The distance traveled by the body before it stops momentarily is

9 m

18 m

54 m

36 m

Medium

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A driver takes $0.20 s$ to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of $54 k m / h$ and the brakes causes a deceleration of $6.0 m / s^{2}$ , find the distance traveled by the car after he sees the need to put the brakes on.

18.63 m

20 m

26.85 m

27.67 m

Medium

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A ball dropped from one metre above the top of a window, crosses the window in $t_{1} s$ . If the same ball is dropped from $2 m$ above the top of the same window, time taken by it to cross the window is $t_{2} s$ . Then

t_{2} = t_{1}

t_{2} = 2 t_{1}

t_{2} > t_{1}

t_{2} < t_{1}

Medium

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A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity $4 \times 1 0^{3}$ $k m / s$ at a distance of $2 c m$ , the time required to reach the given velocity is :

1 0^{- 3} s

1 0^{- 6} s

1 0^{- 8} s

1 0^{- 5} s

Medium

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From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If $V_{A}$ and $V_{B}$ are their respective velocities on reaching the ground then :

V_{B} < V_{A}

V_{A} = V_{B}

V_{A} < V_{B}

Their velocities depends on their masses

Medium

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It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minutes to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator?

30 s e c

45 s e c

40 s e c

35 s e c

Medium

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The average velocity of a body moving with uniform acceleration after travelling a distance of $3.06 m$ is $0.34 m / s$ . The change in velocity of the body is $0.18 m / s$ . During this time, its acceleration is

0.01 m / s^{2}

0.02 m / s^{2}

0.03 m / s^{2}

0.04 m / s^{2}

Medium

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A body travels $200 m$ in the first two second and $220 m$ in the next four second. The velocity at the end of the seventh second from the start will be

m s^{- 1}

m s^{- 1}

220

m s^{- 1}

m s^{- 1}

Medium

View solution>

Motion In A Straight Line

Physics 8636 Views

Two cars are travelling towards each other on a straight road at velocities 15 m/s and 16 m/s respectively. When they are 150 m apart, both the drivers apply the brakes and the cars decelerate at 3 m/s2 and 4 m/s2 until they stop. Separation between the cars when they come to rest is :

The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:

A driver takes 0.20s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54km/h and the brakes causes a deceleration of 6.0 m/s2, find the distance traveled by the car after he sees the need to put the brakes on.

A ball dropped from one metre above the top of a window, crosses the window in t1​s . If the same ball is dropped from 2m above the top of the same window, time taken by it to cross the window is t2​s . Then

A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity 4×103 km/s at a distance of 2cm, the time required to reach the given velocity is :

From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If VA​ and VB​ are their respective velocities on reaching the ground then :

It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minutes to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator?

The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 m/s. The change in velocity of the body is 0.18 m/s. During this time, its acceleration is

A body travels 200m in the first two second and 220m in the next four second. The velocity at the end of the seventh second from the start will be

Physics
8636 Views

A ball dropped from one metre above the top of a window, crosses the window in $t_{1} s$ . If the same ball is dropped from $2 m$ above the top of the same window, time taken by it to cross the window is $t_{2} s$ . Then

A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity $4 \times 1 0^{3}$ $k m / s$ at a distance of $2 c m$ , the time required to reach the given velocity is :

From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If $V_{A}$ and $V_{B}$ are their respective velocities on reaching the ground then :

The average velocity of a body moving with uniform acceleration after travelling a distance of $3.06 m$ is $0.34 m / s$ . The change in velocity of the body is $0.18 m / s$ . During this time, its acceleration is

A body travels $200 m$ in the first two second and $220 m$ in the next four second. The velocity at the end of the seventh second from the start will be