9th Chapter Motion solved questions part -3

CBSE SCIENCE
Question: The distance (S) in metres travelled by a particle is related to time(t) in seconds by the equation of motion S = 5 t2. What is the initial velocity of the particle?
nswer:

 


S = 0 + 5 t2

Initial velocity = 0 [because t is not 0]

Question 22

Question: What conclusions do you draw about the nature of motion of the body from the following velocity-time graph?





Answer: a) The object is moving with uniform acceleration (since the graph is a straight line. Straight line graph means change in velocity in equal intervals of time)

b) The object is having zero acceleration (since the object is moving with uniform velocity)

c) The object is moving with negative acceleration or retardation (since the velocity decreases and finally becomes zero)

d) The object is moving with variable acceleration (since the change in velocity is not equal)

Question: State the type of motion represented by each of the following graphs:



Answer: (i) Body is moving with uniform velocity

(ii) Body is stationary ( at rest )

(iii) Body is moving with uniform retardation (velocity is decreasing uniformly)

(iv) Body is having variable acceleration

(v) Body is travelling with a uniform acceleration but not starting from rest

(vi) Body is moving with uniform velocity

(vii) Body is moving with variable acceleration

(viii) Body is moving with variable acceleration

Question: The figure below gives the v-t graph of a car. What is the distance covered by the car when it is moving with a uniform speed of 15 m/s.



Answer: The distance covered by the car when it is moving with a uniform speed of 15 m/s = area of the rectangle ABDE

Question: Figure shows the velocity time graph of the motion of a body.



(a) State the type of motion in each of the following cases:

(i) OA

(ii) AB

(iii) CD

(b) What is the maximum velocity reached by the body?

(c) State the interval during which the body is moving with a uniform velocity

(d) Calculate the acceleration in the first 3 seconds.

(e) Calculate the retardation.

(f) Calculate the distance travelled in the first 5 seconds.

Answer: (a)

(i) OA uniform acceleration

(ii) AB constant velocity ( no acceleration )

(iii) CD uniform retardation

(b) Maximum velocity attained by the body= 40 m/s.

(c) The body has uniform velocity from t = 3 s to t = 5 s.

(d) Acceleration in the first 3 seconds = slope of the line OA

=  = 10 m/s2

(e) Retardation = slope of the line CD

  

(f) Distance travelled in the first 5 seconds

= Area of the triangle OAG + Area of the rectangle ABFG

    = 45 + 60 = 105 m
Question: A body travelling with a velocity of 200 ms-1 is brought to rest in 10 s. Calculate the retardation.
Answer: u = 200 m/s

v = u + at              0 = 200 + ( a x 10 )                  10 a = -200          a = =

( -ve sign shows retardation )
Question: A body travelling with a velocity of 120 ms-1 accelerates uniformly at the rate of 20 m s-2 for a period of 40 s. Calculate the velocity and the distance travelled in 40 s.
Answer: Given u = 120 m/s,     a = 20 m/s2           t = 40 s
Velocity v = u + at    = 120 + ( 20 x 40 ) = 120 + 800 

v = 920 m/s 

Distance travelled, S =   =  = 4800 + 16000 S = 20800 m

Question: A car starting from rest acquires a velocity of 36 km/h in 5 seconds. Calculate:

(i) its acceleration and (ii) the distance moved by it.

Answer: Initial velocity = 0        Final velocity v = 36 km/h = 10 m s-1

Acceleration (a)   = 2 m s-2 ,   

v2 - u2 = 2aS
Distance  = S

      = 25 m

(i) Acceleration = 2 m s-2

(ii) Distance = 25 m

Question: A body moving with an initial velocity of 36 km/h accelerates uniformly at the rate of 5 m/s2 for 20 seconds. Calculate the total distance travelled in 20 s and the final velocity.

Answer: u = 36 km/h

= 36 x = 10 m s-1

a = 5 m/s2            t = 20 s.
S = ut + at2        

= 200 + 1000   = 1200 m    = 1.2 km

v = u + at = 10 + 5 x 20

= 10 + 100 = 110 m s-1

= 396 km/h
Distance travelled = 1.2 km, final velocity = 396 km/h.
Question: A car travels at a uniform velocity of 60 m s-1 for 10 s and is brought to rest in 5 s. Calculate the retardation of the car and the distance travelled in 15 s.
Answer: Uniform velocity = 60 m s-1  Time = 10 s.

distance travelled in 10 s = velocity x time= 60 x 10 = 600 m.

v = 0, u = 60 m s-1, t = 5 s 

   = -12 m s-2
Retardation = 12 m s-2
Since v2 - u2 = 2aS
02 - 602 = 2 x -12 x S


The car travels 150 m in 5 s 

Distance travelled in 15 s = (600 + 150) = 750 m.