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
>> $Derivation of Equations of ...$

Derivation of Equations of Motion in One Dimension Using Calculus

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

example

Solving problems involving more than one equation of motion

Let the initial velocity of a particle is

5 m / s

and it is accelerated for

2 s

with an acceleration of

1 m / s^{2}

. The distance covered by the particle during this motion can be estimated by:

v = u + a t

v = 5 + 1 \times 2^{2}

v = 9 m / s

Now

9^{2} = 5^{2} + 2 \times 1 \times s

s = 28 m

formula

Third equation of motion

Let,
s -displacement of object startig from rest
u - initial velocity
v - final velocity
t - time of travel
Acceleration is defined as the time rate of change of velocity.

a = \frac{d v}{d t}

multiply and divide numerator and denominator of R.H.S. by

d s

a = \frac{d v}{d t} \times \frac{d s}{d s}

\frac{d s}{d t} = v

hence

a = v \frac{d v}{d s}

a d s = v d v

integrating bothsides, we get

\int_{0}^{s} a d s = \int_{u}^{v} v d v

2 a s = v^{2} - u^{2}

v^{2} = u^{2} + 2 a s

......

(3^{r d} e q u a t i o n o f m o t i o n)

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Velocity - Time Relation in Uniform Accelerated Motion

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Position - Time Relation in Uniform Accelerated Motion

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Position - Velocity Relation in Uniform Accelerated Motion

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Displacement in Nth Second

ExampleDefinitionsFormulaes

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Derivation of Equations of Motion (Calculus Method)

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

Hard

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example

Solving problems involving more than one equation of motion

formula

Third equation of motion

REVISE WITH CONCEPTS

Velocity - Time Relation in Uniform Accelerated Motion

Position - Time Relation in Uniform Accelerated Motion

Position - Velocity Relation in Uniform Accelerated Motion

Displacement in Nth Second

Quick Summary With Stories

Derivation of Equation of Motion(Graphical Method)

Derivation of Equation of Motion(Calculus Method)

Problem Based on Second Equation of Motion

A ball is thrown vertically upwards with a velocity of 20ms−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=10ms−2.

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/s2)

A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance s1​ in the first 10 seconds and distance s2​ in the next 10 seconds, then

If a car at rest accelerates uniformly to a speed of 144 kmh−1 in 20 s, then it covers a distance of:

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10ms−1 to 20ms−1 while passing through a distance 135m in t second. The value of t is

The distance traveled by a particle starting from rest and moving with an acceleration 34​ms−2 in the third second is -

A body A starts from rest with an acceleration a1​. After 2 seconds, another body B starts from rest with an acceleration a2​. If they travel equal distances in the 5th second, after the start of A, then the ratio a1​ : a2​ is equal to

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

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?

A ball is thrown vertically upwards with a velocity of $20 m s^{- 1}$ 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 $g = 10 m s^{- 2}$ .

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 $t = 18 s$ . What is the value of $v$ ? (Take $g = 10 m / s^{2}$ )

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

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

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

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

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 $5^{t h}$ second, after the start of $A$ , then the ratio $a_{1}$ : $a_{2}$ is equal to

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