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: 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: 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 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: 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 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: 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

Physics
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Motion In A Straight Line

Physics 725 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

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 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 :

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?

Physics
725 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

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 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 :