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: For a body moving with uniform acceleration \$a\$, initial and final velocities in a time interval \$t\$ are \$u\$ and \$v\$ respectively. Then, its average velocity in the time interval \$t\$ is :

\$\textbf{Hint}\$: Use the equations of motion \$\textbf{Step 1}\$: Given a body is moving with uniform acceleration 'a', initial velocity is u and final velocity is v, time is t. We have to find the average velocity. From the equations of motion we know\$s=ut+\dfrac{1}{2}at^{2}\$and \$v=u+at\$ If an object velocity is increasing at a constant rate then it is called as uniform acceleration. \$\textbf{Step 2}\$:We know Average velocity can be calculated using the formula\$Average \quad velocity= \dfrac{Total \quad displacement}{Total \quad Time}=\dfrac{ut+\dfrac{1}{2}at^{2}}{t}\$\$Average \quad velocity=u+\dfrac{1}{2}at\$From equations of motion \$v=u+at\$ Using this equation we can write \$u=v-at\$Substitute \$u=v-at\$ in the above equation obtained for Average velocitywe get\$Average \quad velocity=v-at+\dfrac{1}{2}at=v-\dfrac{1}{2}at\$So both option B and D are correct.

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 bus starts from rest and moves with a uniform acceleration of \$1\;ms^{-2}\$. A boy \$10\;m\$ behind the bus at the start runs at a constant speed and catches the bus in \$10\;s\$. Speed of the boy is :

Solution 2Distance travelled by bus in \$10s \Rightarrow \dfrac {1}{2} \times 1 \times 10^2 = 50 m\$D \$=\$ Distance boy has to cover \$= 10 + 50 = 60 m\$Speed of boy \$=\dfrac {D}{T} = \dfrac {60}{10} = 6\ m/s\$

Q: A train accelerates from rest at a constant rate \$a_{1}\$ for distance \$S_{1}\$ and time \$t_{1}\$. After that it decelerates to rest at a constant rate \$a_{2}\$ for distance \$S_{2}\$ at time \$t_{2}\$. Then the correct relation among the following is

Given that the particle was at rest at \$t\ =\ 0\$\$s_1 = u_1t + \frac {1}{2} a_1t_1^2 .... (1)\$\$\Rightarrow v_1 = u_1 + a_1t_1 .... (2)\$\$v_1 \rightarrow \$ final speed after \$t_1\$Now, final speed after this is \$ v_2\$And \$v_1 = v_2\$\$\Rightarrow v_2 = u_2 - a_2t_2\$\$\Rightarrow v_1 = a_2t_2 .... (3)\$\$\Rightarrow v_2^2 = u_2^2 - 2a_2s_2\$\$\Rightarrow v_1^2 = 2a_2s_2\$From (2), \$a_1^2t_1^2 = 2a_2s_2 \$\$\Rightarrow s_2 = \dfrac {a_1^2t_1^2}{2a_2}\$\$\Rightarrow s_2 = \dfrac {a_1 (2s_a)}{2\frac {a}{2}a_2}\$\$\Rightarrow \dfrac {eq^2 (2)}{eq^n (3)} = \dfrac {a_1t_1}{a_2t_2} = 1\$\$\Rightarrow \dfrac {a_1}{a_2} = \dfrac {t_2}{t_1}\$\$\Rightarrow \dfrac {a_1}{a_2} = \dfrac {t_2}{t_1} = \dfrac {s_2}{s_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: 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\$

Physics
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Motion In A Straight Line

Physics 4727 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.

For a body moving with uniform acceleration a, initial and final velocities in a time interval t are u and v respectively. Then, its average velocity in the time interval t is :

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 bus starts from rest and moves with a uniform acceleration of 1ms−2. A boy 10m behind the bus at the start runs at a constant speed and catches the bus in 10s. Speed of the boy is :

A train accelerates from rest at a constant rate a1​ for distance S1​ and time t1​. After that it decelerates to rest at a constant rate a2​ for distance S2​ at time t2​. Then the correct relation among the following is

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 :

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

Physics
4727 Views

For a body moving with uniform acceleration $a$ , initial and final velocities in a time interval $t$ are $u$ and $v$ respectively. Then, its average velocity in the time interval $t$ is :

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 bus starts from rest and moves with a uniform acceleration of $1 m s^{- 2}$ . A boy $10 m$ behind the bus at the start runs at a constant speed and catches the bus in $10 s$ . Speed of the boy is :

A train accelerates from rest at a constant rate $a_{1}$ for distance $S_{1}$ and time $t_{1}$ . After that it decelerates to rest at a constant rate $a_{2}$ for distance $S_{2}$ at time $t_{2}$ . Then the correct relation among the following is

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