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**Q1. Give the difference between distance and displacement. Ans.**

Distance | Displacement |

1. It is the total path covered by an object. | 1. It is the shortest path measured from initial position to final position. |

2. It does not need direction. | 2. It needs direction. |

3. It cannot be ‘0’. | 3. It can be ‘0’. |

**Q2. Give the difference between speed and velocity. Ans.**

Speed | Velocity |

1. It is the distance travelled by an object per unit time. | 1. It is the displacement of the body per unit time. |

2. It is scalar quantity, direction not required. | 2. It is a vector quantity, direction is required. |

3. Its unit is m/s and is always positive. | 3. Its unit is m/s and can be negative. |

**Q3. Give the difference between acceleration and deceleration. Ans.**

Acceleration | Deceleration |

1. It is change in velocity per unit time. It is always positive. | 1. It is change in velocity per unit time. It is always negative. |

2. In this case, the velocity keeps increasing. | 2. In this case the increased velocity starts decreasing. |

3. u < v, initial velocity is less than final velocity u = 0. | 3. u > v, final velocity is less than the initial velocity v = 0. |

**Q4. Differentiate between velocity and acceleration. Ans.**

Velocity | Acceleration |

1. The rate of change of position of body is called velocity. | 1. It is the rate of change of velocity. |

2. S.I. unit is m/s. | 2. S.I. unit is m/s2. |

**Q5. With the help of graph, show the uniform acceleration and uniform retardation of a body. Ans.**

OA − shows uniform acceleration.

CB − shows uniform retardation.

**Q6. What conclusion do you draw about acceleration of the particle in motion from given velocity-time graph?**

**Ans. **(a) No acceleration

(b) Uniform acceleration

**Q7. A car accelerates uniformly from 20 km/h to 35 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time. Ans.**

u= 20km/h = 5.5m/s

v= 35km/h = 9.72 m/s

t=5s

**(ii) The distance covered by the car**

**Q8. The odometer of a car reads 1800 km at the start of a trip and 2400 km at the end of the trip. If the trip took 10 h. Calculate the average speed of the car in km/h and m/s. Sol. **Distance covered by the car

s = 2400 − 1800 = 600 km.

Trip time = 10 hr

Average speed = ?

The average speed of the car in km/h = 60 km/h and in m/s is 16.7 m/s.

**Q9. An object travels 16 m in 4 s and then another 16 m in 3 s. What is the average speed of the object? Ans. **Total distance travelled by the object = 16 m + 16 m = 32 m

Total time taken = 4s +3s = 7s

∴ Average speed = Total distance travelled/ Total time taken

= 32/7 = 4.57 m/s

∴ The average speed of the object is 4.57 m/s.

**Q10. It took 2 s after lightning for the sound of thunder to reach you. How far did the lightning struck? Ans. **Speed of sound is 346 m/s in air.

Time = 2 s

∴ Distance = ?

s = (d/t)

∴ d = s × t = 346 × 2 = 692 m.

∴ The lightning struck 692 m away.

**Q11. Find the type of motion in the following case : (a) A car travelling along a straight road, changes its speed. (b) An athlete running a 100 m race. (c) Moon revolving around the earth. (d) An ant moving on the floor. Ans.** (a) Non-uniform motion

(b) Uniform motion

(c) Uniform circular motion

(d) Non-uniform motion

**Q12. What conclusion do you draw about acceleration of the particle in motion from given velocity-time graphs.**

**Ans**. (a) No acceleration (b) Uniform acceleration (c) Non-uniform acceleration

**Q13. Derive the equation for velocity-time relation (v = u + at) by graphical method. Ans.** Let the initial velocity of a body be u = OA = CD.

The final velocity of a body be v = OE = CB

Time t = OC = AD

AD is parallel to OC

BC = BD + DC

= BD + OA

∴ v = BD + u

∴ BD = v − u ...(1)

In velocity-time graph, slope gives acceleration.

∴ BD = at ...(2)

Substitute (2) in (1) we get.

BD = v − u

at = v − u

∴ v = u + at

**Q14. How can we get speed from distance-time graph? Ans.** Let us assume that an object moves with a uniform speed. The distance-time graph will be a straight line.

let d_{1} be the distance covered in time t_{1}

Let d_{2} be the distance covered in time t_{2}

∴ d_{2} - d_{1} will be the distance covered in t_{2}-t_{1}

Conclusion:

To find the speed from distance-time fraph we can get ots slope which equal to the speed of the object

**Q15. Represent a graph that shows the following: (a) Uniform speed (b) Non-uniform speed (c) Stationary object Ans.** Time is taken on x-axis as it is independent quantity and all dependent quantities are taken along y-axis.

**Q16. Draw graph to show the following: (a) Uniform acceleration (b) Non-uniform acceleration (c) Uniform motion Ans. (a)**

Uniform acceleration - A body travels equal distance in equal intervals of time.

Non-uniform acceleration - A body travels unequal distance in equal interval of time.

Uniform motion or zero accelation - A body moves with constant speed in same line.

**Q17. How can you get the distance travelled by on object from its speed-time graph? Ans. **Suppose an object is moving with a same speed v, the distance covered by this object when it was at (point A)

t_{1} to t_{2} is :

AB = t_{2 }- t_{1}

AD = BC = v

∵ v = distance/time

∴ distance = v(t_{2 }- t_{1)}

_{ }= AD.AB

∴ s = area of rectangle ABCD

∴ To find the distance travelled by a body from its speedtime graph, we need to find the area enclosed by the graph.

**Q18. Define acceleration and give its SI unit. When is acceleration of a body negative? Give two examples of situations in which acceleration of the body is negative. Ans. **The change in velocity per unit time is called acceleration. Its unit is m/s

When the velocity of the body decreases it is negative, for e.g., a moving car applies breaks to stop, it is called deceleration. When a body falling due to the gravity, free fall is also considered to be negative –9.8m/s

**Q19. Distinguish between uniform motion and non- uniform motion. Ans.**

**Non-uniform motion: **When a body travels unequal distance in equal intervals of time it is called non-uniform motion.

**Q20. The below data shows the speedometer readings of a car. Calculate the acceleration of the car.**

**Ans.**

**Q21. Velocity-time graph for the motion of an object in a straight path is a straight line parallel to the time axis. Comment on any two analysis from such a graph. Ans. **The nature of the motion is uniform.

The object is not accelerating.

**Q22. Draw the shape of the distance-time graph for uniform and non-uniform motion of object. Ans.**

**Q23. A bus of starting from rest moves with uniform acceleration of 0.5 ms ^{–2} for 2 minutes. Find (a) the speed acquired. (b) the distance travelled. Ans. (a)**

**(b)**

**Q24. A car accelerates uniformly from 50 km/h to 85 km/h in 10 seconds. Calculate (a) acceleration in m/s ^{2}. (b) distance covered by the car in metres during the given time of interval.**

Ans. (a) a = 85 – 50 = 35 km/h =

a = 9.73m/s

50km/h = 13.89 m/s

85km/h = 23.62 m/s

(b) There are two ways to find the distance covered by the car in t = 10 seconds.

**Q25. A vehicle decelerates from 68 ms ^{–1} to 16 ms^{–1} in 8 sec. Calculate**

(a) the deceleration and

(b) the distance covered by the car in that time.

Ans. (a)

**(b) Distance covered **

= 68 × 8 – 0.5 × 6.5 × 8 × 8 = 544 – 208 = 336 m

** Q26. (a) Differentiate between speed and velocity. (b) When is a body said to have uniform velocity? (c) How can we describe the position of an object? Illustrate with suitable example. Ans.** (a) Speed is distance traveled per unit time and velocity is displacement per unit time.

(b) Uniform velocity: when a body travels equal distance in equal intervals of time.

(c) The position of an object is described by the displacement shown by an object from its original position. For e.g., a boy travels in a circular track from a start point and reaches the finish point which is the same point. Hence the displacement is zero.

**Q27. Define the term displacement. Is it a vector quantity or a scalar quantity? Ans.** The distance moved by a body from its mean position is called displacement. It is a vector quantity.

**Q28. What is circular motion? Is circular motion an acceleration motion? Ans.** Uniform circular motion can be described as the motion of an object in a circle at a constant speed. Yes, as an object moves in a circle, it constantly changes its direction.

**Q29. A car is travelling with a speed of 90km/h. The driver applied the brakes and retards the car uniformly. The car is stopped in 10sec. Find: (a) The acceleration of car. (b) Distance before it stops after applying breaks. Ans.**

**Q30. Give two examples of a situation when the displacement can be zero. Ans. **(a) The displacement can be zero if the start point of the object is same as the end point.

(b) The body reaching the same point after travelling a particular distance. A body in circular motion.

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