Motion in One Dimension **Exe-2C** Equations of Motion Very Short Answer Type for Class-9 ICSE Concise Physics. There is the solutions of Very short Answer and Long answer type Questions of your latest textbook which is applicable in 2023-24 academic session**. **Visit official Website CISCE for detail information about ICSE Board Class-9.

**Motion in One Dimension Exe-2 C Equations of Motion Very Short Answer **

**(ICSE Class – 9 Physics Concise Selina Publishers)**

Board | ICSE |

Class | 9 |

Subject | Physics |

Writer / Publication | Concise selina Publishers |

Chapter-2 | Motion in One Dimension |

Exe – 2 C | Equations of Motion |

Topics | Solution of Exe-2(C) Very Short and Long Answer Type |

Academic Session | 2023-2024 |

**Exe-2C Equations of Motion Very Short Answer Type**

**Ch-2 Motion in One Dimension Physics Class-9 ICSE Concise**

**Page 60**

**Question 1. **Write three equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).

**Answer:**

Three equations of a uniformly accelerated motion are

v = u + at

s = ut + (1/2)at^{2}

v^{2} = u^{2} + 2as

**Question 2. **Write an expression for the distance S covered in time t by a body which is initially at rest and starts moving with a constant acceleration a.

**Answer:**

Distance = s, time = t, initial velocity u = 0 and acceleration = a.

Using the second equation of motion and substituting the above values we get,

**s = ut + (1/2) at ^{2} **

**Exe-2C Equations of Motion Long Answer Type**

**Motion in One Dimension Exe-2C Very Short Class-9 ICSE**

**Page 60**

**Question 1. **Derive the following equations for uniformly accelerated motion:

(i) v = u + at

(ii) s = ut + (1/2) at^{2}

(iii) v^{2} = u^{2} + 2aS

where the symbols have their usual meanings.

**Answer:**

**Derivation of equations of motion**

**First equation of motion:**

Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its initial velocity, and at t = t, let v be its final velocity.

Acceleration = *Change* *in* *velocity/Time* a = (*v *– *u)/t* at = v – u**v = u+ at … First equation of motion.**

**Second equation of motion:**

Average velocity = *Total* *distance* *traveled/Total* *time* *taken*Average velocity = *s/t *…(1)

Average velocity can be written as (*u*+*v)/*2 Average velocity = (*u*+*v)/*2 …(2)

From equations (1) and (2) *s/t* = (*u*+*v)/*2 …(3)

The first equation of motion is v = u + at.

Substituting the value of v in equation (3), we get

*s/t* = (*u *+ *u *+ *at*)/2 s = (2*u *+ *at*) *t/*2 = 2*ut *+ *at*^{2}/2 = 2*ut/*2 + *at*^{2}/2

**s = ut + (1/2) at ^{2} …Second equation of motion**.

**Third equation of motion: **

The first equation of motion is v = u + at. v – u = at … (1)

Average velocity = *s/t* …(2)

Average velocity =(*u*+*v)/*2 …(3)

From equation (2) and equation (3) we get,

*(u *+ *v)/*2 = *s/t* …(4)

Multiplying eq (1) and eq (4) we get,

(v – u)(v + u) = at × (2*s/t)* (v – u)(v + u) = 2as

[We make the use of the identity a^{2} – b^{2} = (a + b) (a – b)]

**v ^{2} – u^{2} = 2as …Third equation of motion.**

— : End of Motion in One Dimension Exe-2C Equations of Motion Very Short and Long Answer Type Solutions :–

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