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**Short Answer Type Question**

**Q.1. The brakes applied to a car produce an acceleration of 6 ms ^{-2} in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time. [CBSE 2016]Ans. **Acceleration, a = - 6 ms

Final velocity, v = 0

Time, t = 2 s

Initial velocity, u = v - at = 0 - (-6) (2)

= 12 ms

Distance travelled,

= 12 m

(i) A to B

(ii) B to C

(iii) C to D

Ans.

(ii) Speed from B to C = slope BC = zero

(iii) Speed from C to D = slope CD = 2 cm/s

(b) The slope of the line on a position-time graph reveals information about an object’s velocity. What conclusion can you draw regarding the motion of an object, if the position-time graph is a:

(i) horizontal line parallel to time axis.

(ii) straight line at 45° to time axis.

(iii) curve [CBSE 2016]

Ans.

v = 60 km/h =

Time taken,

(b) (i) The object is at rest

(ii) The object is moving at constant velocity.

(iii) The motion of the object is non- uniform.

Ans.

Let body moves 20 m from B to C.

Distance = AB + BC

= 15 m + 20 m = 35 m

Displacement =

(a) What is its average acceleration?

(b) How far does it travel in that time?

Ans.

Ans.

Ans.

Time 5 h = 5 x 3600 s

Speed =

Ans.

Distance = vt = 30 x 30 = 900 m

Ans.

∴ acceleration

and distance covered

After applying the brakes in second case initial velocity u

∴ acceleration

and distance covered

Total distance covered by Anil s = s

Ans.

∴ acceleration

For subsequent motion, initial velocity u' = 12 m s

∴

(a) the slope of the distance-time graph.

(b) the area under velocity-time graph.

(c) the slope of the velocity-time graph. [CBSE 2013, 2016]

Ans.

(b) The area under velocity-time graph gives the magnitude of displacement (or the distance) covered by the object.

(c) The slope of velocity-time graph gives the acceleration of the moving object.

Ans.

The motion of an object is said to be non-uniform if it covers unequal displacements in equal intervals of time or equal displacements in unequal intervals of time.

(a) If a moving object is to have a uniform speed then the shape of the path covered by it can be either a straight line or a curve.

(b) If a moving object is to have uniform velocity then its path must be a straight line path along a given direction.

(i) m/s

(ii) m/s

Ans.

Ans.

∴ Acceleration

Ans.

Ans.

Ans.

For motorcycle B :

Speed-time graphs for motorcycles A arid B have been plotted in Fig. 8.27.

Distance covered by motorcycle A

s

and distance covered by motorcycle B

s

Thus, both motorcycles travelled equal distances before coming to a stop.

(i) State the type of motion shown by a freely falling stone.

(ii) When a stone is thrown vertically upwards its velocity is continuously decreases. Why?

(iii) Give an example of a motion in which average velocity is zero, but the average speed is not zero. [CBSE 2015]

Ans.

(ii) When a stone is thrown vertically upwards, a constant acceleration due to gravity acts on it in the downward direction. It causes a retardation. So, velocity of stone continuously goes on decreasing.

(iii) Let a person goes from station A to station B and then returns back to station A covering a finite distance in certain time. Here average speed of person is finite. However, his average velocity is zero because there is no net displacement of the person.

Ans.

Ans.

**Long Answer Type Question**

**Q.1. The speed-time graph of a body is as shown in Fig. 8.34:****(a) What type of motion is represented by OA?****(b) What type of motion is represented by AB?****(c) Calculate the retardation of the body.****(d) Calculate the distance travelled by the body from O to C.****(e) What is the average velocity of the body for its entire journey? [CBSE 2016]****Ans. **(a) OA represents uniformly accelerated motion.

(b) AB represents uniform motion without any acceleration.

(c) BC represents the retarded motion, in which magnitude of retardation

(d) Distance travelled by the body from O to C, s = area of the trapezium OABC

(e) Average velocity of the body for entire journey = **Q.2. An object starts linear motion with a velocity u and under uniform acceleration a it acquires a velocity v in time t. Draw velocity-time graph. From this graph obtain the following equations: [CBSE 2015, 2016](a) v = u + at,(b) s = ut + 1/2 at**The velocity-time graph has been shown in Fig. 8.30.

At t = 0, the initial velocity is u (at point A) and then increases to v (at point B) in time t.

(a) Draw perpendicular lines BC and BE from point B on the time and velocity axes respectively, so that OA = u, OE =BC = v.

Also draw a line AD parallel to time axis, so that OC = AD = t.

Then change in velocity in time t = BC - OA = BC - CD = BD

But BC = v and CD = OA = u. Hence, BD = v - u

From the velocity-time graph, the acceleration of the object is given by

∴ at = v - u ⇒ v = u + at

(b) According to velocity-time graph the total distance covered by the object is obtained by the area under the graph.

Hence, distance s travelled by the object = area of the trapezium OABC

= area of rectangle OADC + area of triangle ADB

But OA = u, OC = t, AD = OC = t and BD = change in velocity = v - u = at

∴

Ans.

s = area of the trapezium OABC

...(i)

From velocity-time relation v = u + at, we have

...(ii)

Substituting value of time t from equation (ii) into (i), we get

⇒ 2as = v

which is the requisite equation.

(а) Identify the kind of motion of the object represented by lines OA and BC.

(b) With what velocity the object is moving at t = 8 seconds?

(c) Calculate the acceleration of the object in the following cases:

(i) Between the third and tenth second.

(ii) During the last two seconds. [CBSE 2015]

Ans.

(b) At time t = 8 s, the object is moving with a velocity of 4.6 m s

(c) (i) Between 3rd and 10th second, object is moving with a constant velocity of 4.6 m s

(ii) During the last two seconds, velocity of object falls from u = 4.6 m s

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