MCQ: Equations of Motion - Class 9 MCQ

Given:
u = 36 km/h = 36 × (5/18) = 10 m/s
Time of reaction, t = 0.9 s
Deceleration, a = -5 m/s²
v = 0 m/s

Let s1 be the distance covered by the man when the object is seen by him.

We have, u = s1 / t
Therefore,
s1 = u × t
s1 = 10 × 0.9
s1 = 9 m

When he applies brakes and decelerates at the rate of 5 m/s², the distance covered by him is s2.

We have,
v² = u² + 2as2
0 = u² - 2as2 (since v = 0)
s2 = u² / 2a
s2 = 10² / (2 × 5)
s2 = 10 m

So, total distance covered s = s1 + s2
s = 9 + 10 = 19 m

Hence, the man covers a total distance of 19 m before he stops.

Topic in NCERT: Graphical Representation of Motion

Line in NCERT: error Occcured while getting response from embedding

Velocity acquired by the two bodies falling from rest through a distance his given as v = 

Given that :

Initial velocity, u = 0.2 m/s. 

Final velocity, v = 0 

Acceleration, a = 0.02 m/s^2.

v = u + at 

0 = 0.2 - 0.02 x t

0.02 x t = 0.2 

t = 0.2 / 0.02 = 10 sec.

Thus, the body will take 10 seconds to stop.

Topic in NCERT: Uniform Circular Motion

Line in NCERT: "If you carefully note, on being released the stone moves along a straight line tangential to the circular path."

Yes this is because we need to observe the motion of object or rest of object from any point called reference point.

Topic in NCERT: Describing Motion

Line in NCERT: "We describe the location of an object by specifying a reference point."

v = u + at

Where:

• v is the final velocity
• u is the initial velocity
• a is the acceleration (since the train is slowing down, this will be negative)
• t is the time

Given:

• Initial velocity u = 80 km/h = (80 × 1000) / 3600 = 22.22 m/s
• Final velocity v = 20 km/h = (20 × 1000) / 3600 = 5.56 m/s
• Deceleration a = -2 m/s²

Now, substitute the values into the equation:
5.56 = 22.22 + (-2)t

Rearranging for t:
5.56 - 22.22 = -2t
-16.66 = -2t
t = 16.66 / 2 = 8.33 seconds

Consider a velocity - time graph for a uniformly accelerated body starting from rest is represented as follows.

u = velocity at time t1 
v =  velocity at time t2 
If acceleration is represented as a, then, acceleration is defined as the rate of change in velocity. 
⇒  a = v-u / t2 - t1 
⇒ a = v-u /t 
Or,v-u = at 
⇒ v= u + at 

Topic in NCERT: VELOCITY-TIME GRAPHS

Line in NCERT: "The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time."

S = ut + (1/2)at²

Where:

• S is the distance covered,
• u is the initial velocity (since the car is starting from rest, u = 0),
• a is the acceleration,
• t is the time.

Given:

• u = 0 m/s,
• a = 6 m/s²,
• t = 10 s.

Substituting the values into the equation:
S = 0 × 10 + (1/2) × 6 × (10)²
S = (1/2) × 6 × 100 = 3 × 100 = 300 m

So, the car will cover 300 meters.

The correct answer is:
Answer: A
300 m

This is because in circular motion body is continuously changing its speed and acceleration is the change of velocity.

Topic in NCERT: UNIFORM CIRCULAR MOTION

Line in NCERT: "When an object moves in a circular path with uniform speed, its motion is called uniform circular motion."

Answer: A. s = ut + (1/2)at²