A particle starts from the origin at t = 0 s with a velocity of 10.0 \$\hat{j}\$ m/s and moves in the x-y plane with a constant acceleration of \$\left ( 8.0\hat{i}\,+\,2.0 \hat{j}\right )\$ m/\$s^2\$.(a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?(b)What is the speed of the particle at the time ?

Given:The initial velocity of particle is \$\vec u = 10 \hat j m/s\$The acceleration of the particle is \$\vec a = 8 \hat i +2 \hat j\$(a)The position of the particle at any instant is given by:\$\vec s = \vec ut + \dfrac{1}{2}{\vec at}^2\$\$x \hat i + y \hat j = 4t^2 \hat i + (10t + t^2)\hat j\$Comparing the X and Y component of the position:\$x=4t^2\$\$x=16 \implies t=2 s\$\$y=10t+t^2\$At t=2, \$y=24 m\$(b)The speed of the particle at any instant is given by:\$\displaystyle {\vec {v}=\vec {u}+\vec a t}\$\$ {\vec {v} = 10\hat j +({8\hat i+2\hat j})2} \$\${\vec{v} = 14\hat j +16\hat i}\$\$|\vec{v}| = \sqrt {14^2+16^2} = 21.26 m/s\$

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>>Class 11
>>Physics
>>Motion in a Straight Line
>>Uniformly Accelerated Motion
>> $Position - Time Relation in...$

Position - Time Relation in Uniform Accelerated Motion

Q: A ball is thrown vertically upwards with a velocity of \${20 ms^{-1}}\$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take \${g= 10 ms^{-2}}\$.

Time to reach maximum height can be obtained from \$v=u+at\$\$0=20+(-10)t\$\$t=2 s\$\$s=ut+0.5at^2=20(2)+0.5(-10)(2)^2= 20 m\$Thus, total distance for maximum height is 45 m\$s=ut+0.5at^2\$\$45=0+0.5(10)(t')^2\$\$t'=3 s\$Total time= 3+2= 5s

Q: A ball is dropped from a high rise platform at \$t=0\$ starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed \$v\$. The two balls meet at \$t=18s\$. What is the value of \$v\$? (Take \$g=10m/{s}^{2}\$)

The distance traveled by the first ball in 18s is \$h=\dfrac{1}{2}gt^2=\dfrac{1}{2}\times 10\times (18)^2=1620 m\$Now to meet second ball has to same distance in\$(18-6)=12s\$So for second ball,\$h=vt+\dfrac{1}{2}gt^2\$\$1620=v\times 12+ \dfrac{1}{2}\times 10\times (12)^2\$\$v=\dfrac{1620-720}{12}=75 m/s\$

Q: A particle experiences constant acceleration for \$20\$ seconds after starting from rest. If it travels a distance \${ s }_{ 1 }\$ in the first \$10\$ seconds and distance \${ s }_{ 2 }\$ in the next 10 seconds, then

Let \$a\$ be the constant acceleration of the particle. Then\$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }\$ or \${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times { \left( 10 \right) }\displaystyle^{ 2 }=50a\$and \${ s }_{ 2 }=\left[ 0+\displaystyle\dfrac { 1 }{ 2 } a{ \left( 20 \right) }^{ 2 } \right] -50a=150a\$\$\therefore \quad { s }_{ 2 }=3{ s }_{ 1 }\$ Alternatively:Let \$a\$ be constant acceleration and \$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }\$, then \${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a\$ Velocity after 10 sec. is \$v=0+10a\$ So, \${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }\$ Let \$a\$ be constant acceleration, using \$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }\$ so distance coreved in first 10 seconds \${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a\$ Velocity after 10 sec. is \$v=0+10a\$ So, distance covered in next 10 seconds \${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }\$

Q: If a car at rest accelerates uniformly to a speed of 144 km\$h^{-1}\$ in 20 s, then it covers a distance of:

Given\$u=0\$\$v=144kmph=144\times\frac{5}{18}=40 m/s\$\$t=20sec\$We know,\$v=u+at\$\$40=a\times20\$\$a=2m/s^2\$Now,\$s=ut+\frac{1}{2}at^2\$\$s=\frac{1}{2}at^2=\frac{1}{2}\times2\times20^2=400m\$

Q: A particle moves in a straight line with a constant acceleration. It changes its velocity from \$ 10 ms^{-1} \ to \ 20 ms^{-1}\$ while passing through a distance \$135m\$ in \$t\$ second. The value of t is

\$u=10m/s,v=20m/s,x=130m\$\$a=\cfrac{v-u}{t}=\cfrac{20-10}{t}=\cfrac{10}{t}\$\$s=ut+\cfrac{1}{2}at^2\\\implies 135 =10t+\cfrac{1}{2}\cfrac{10}{t}t^2\\\implies t=9s\$

Q: The distance traveled by a particle starting from rest and moving with an acceleration \$\dfrac{4}{3}ms^{-2}\$ in the third second is -

Distance covered in nth second,\$S_{n}=u+\dfrac{1}{2}a(2n-1)\$;\$=0+\dfrac{1}{2}\times\dfrac{4}{3}(2\times3-1)\$\$=\dfrac{1}{2}\times\dfrac{4}{3}\times 5=\dfrac{10}{3} m\$

Q: A body \$A\$ starts from rest with an acceleration \$a_1\$. After \$2\$ seconds, another body \$B\$ starts from rest with an acceleration \$a_2\$. If they travel equal distances in the \$5^{th}\$ second, after the start of \$A\$, then the ratio \$a_1\$ : \$a_2\$ is equal to

Initial speed \$u = 0\$Distance covered in nth second after starting from rest \$S = \dfrac{1}{2}(2n-1)\$For A : \$n = 5\$So, \$S_A = \dfrac{a_1}{2}(2\times 5-1) = \dfrac{9}{2}a_1\$For B : \$n = 3\$So, \$S_B = \dfrac{a_2}{2}(2\times 3-1) = \dfrac{5}{2}a_2\$But \$S_A=S_B\$Or \$\dfrac{9}{2}a_1 = \dfrac{5}{2}a_2\$\$\implies \ a_1:a_2 = 5:9\$

Q: A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m.If the ball is in contact with the floor for 0.01 sec, then average acceleration during contact is

Let u be the velocity with which the ball hits the ground, then\$\displaystyle { u }^{ 2 }=2gh\$\$\displaystyle =2\times 9.8\times 10=196\$\$\displaystyle \therefore \quad u=14 m/sec\$If v be the velocity with which it rebounds, then\$\displaystyle { v }^{ 2 }=2\times 9.8\times 2.5=49\$\$\displaystyle \Rightarrow v=7m/sec\$\$\displaystyle \therefore \quad \Delta v=\left( v-u \right) \$\$\displaystyle =\left( 7m/sec \right) -\left( -14m/sec \right) \$\$\displaystyle =21m/sec\$\$\displaystyle \therefore \quad a=\frac { \Delta v }{ \Delta t } =\frac { 21 }{ 0.01 } \$\$\displaystyle =2100{ m }/{ { s }^{ 2 } }\$

Q: A particle starts from rest, accelerates at \$2\ ms^{-2}\$ for \$10\ s\$ and then goes for constant speed for \$30\ s\$ and then deceleration at \$4\ ms^{-2}\$ till stops after next. What is the distance travelled by it?

Displacement in first, \$10\,\sec \$ \${{s}_{1}}=\dfrac{1}{2}{{a}_{1}}{{t}_{1}}^{2}=\dfrac{1}{2}\times 2\times {{10}^{2}}=100\,m\$ Speed achieve in first \$10\,\sec \$ \${{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}}=0+2\times 10=20\,m/s\$ Displacement in \$30\,\sec \$, acceleration is zero. \${{s}_{2}}={{v}_{1}}{{t}_{2}}=20\times 30=600\,m\$ Displacement in last due to deacceleration, \${{a}_{3}}=-4\,m/{{s}^{2}}\$ \$ {{v}^{2}}-{{v}_{1}}^{2}=2as \$ \$ {{s}_{3}}=\dfrac{0-{{20}^{2}}}{2\times (-4)}=50\,m \$ Net displacement \${{S}_{total}}={{s}_{1}}+{{s}_{2}}+{{s}_{3}}=100+600+50=750\,m\$ Total displacement is \$750\,m\$

Physics

formula

Second equation of motion

Distance covered by the object in the given time t is given by the area of the trapezium ABDOE

Let in the given time, t the displacement covered by the moving object = s

The area of trapezium, ABDOE

Displacement (s) = Area of ABD + Area of ADOE

= $\frac{1}{2} \times A B \times A D + A E \times O E$ , $A B = D C = a t$

= $\frac{1}{2} a t \times t + u t$

Hence

$s = u t + \frac{1}{2} a t^{2}$ ............... $(2^{n d} e q u a t i o n o f m o t i o n)$

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