20 Physics -- Kinematics

Q.A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about *A.50km/h toward westB.70km/h toward south-westC.70km/h toward north-westD.zero

Q.A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about *

A.50km/h toward west

B.70km/h toward south-west

C.70km/h toward north-west

D.zero

1 Answers

1101 Views

The magnitude as well as the direction of acceleration of particle is constant. Will its path be definitely rectilinear?

Considering projectile motion, we cannot say that the motion of an object remains rectilinear under constant magnitude of acceleration and its direction.

In a projectile motion, the acceleration is constant and equal to 'g' which is always directed towards the centre of the Earth. However, when it comes to the path followed by the projectile, it is parabolic in nature.

A rescue plane is flying at constant elevation of 1200m with a speed of 430km\hr towards a point directed over a person struggling in water. At what angle of slight should the pilot release a rescu...

Time of flight(T)=(2H/g)

Range(R)=u*T

R=430*1000/3600*(2*1200/9.8)

R=1869.21m

Now,

tan(theta)=H/R=0.64

Theta=32.62

A projectile moves in a parabolic path without air resistance. Is there any point at which acceleration is i. parallel to the velocity? ii. Perpendicular to the velocity?

Since the horizontal component of the velocity u of such a projectile is constant, there is no acceleration in this direction. So the acceleration of the projectile is taking place in the vertical direction. The acceleration is parallel or anti-parallel to the vertical component of u and hence there is no point at which the acceleration is parallel to u. At the maximum height, the vertical component of u is zero but u is horizontal. So the acceleration is perpendicular to u at the maximum...

Is it possible for an object (i) to be slowing down while its acceleration is increasing in magnitude? (ii) to be speeding up while its acceleration decreasing? In each case, explain you reasoning.

The acceleration of an object refers to the change of velocity at a certain rate including both slowing down and speeding up.
(i) So it is possible for the object to be slowing down while its acceleration is increasing in magnitude. In such a case, the object suffers an extreme deceleration and comes to rest rapidly.
(ii) It is also possible for the object to be speeding up while its acceleration is decreasing. In such a case, the acceleration goes on decreasing to become zero and finally the...

Kohli is chasing a cricket ball and both of them are moving on the same path. After they pass mid off, their velocities v vary with time t as shown in the graph above. When will the Kohli's chase end?

1. Figure 3.64 shows the position-time graphs of three cars A, B, and C. On the basis of the graphs, answer the following questions:i) Which car has the highest speed and which is the lowest?ii) ...

Solution

Velocity of car A = slope of x-t of A = 14-6/1.6 = 5km/hs

Velocity of car B = slope of x-t of B =14-0/1.4 = 10km/hs

Velocity of car C = slope of x-t of c = 14-2/1 =12km/hs

A) Car C has the highest speed and car A has the lowest.

B) There is no point where all three line intersect. so, all three car never meet at a same position.

C) When car A passes car C, car B is at 6km.

D) Time interval= 1.2-0.6 = 0.6hr

E) Relative velocity of Car C with respect to car A(v_CA)=V_C-V_A=12-5 =7km/hr

F) R...

2. A boy stretches a stone against the rubber tape of a catapult or gulel' (a device used to detach mangoes from the tree by boys in Indian villages) through a distance of 24 cm before leavi...

Initial velocity(u)= 0m/s

Final velocity(v)= 2.2m/s

Displacement(d)= 24cm=0.24m

Now, v²=u²+2as

(2.2)² = 0²+2a0.24

a = 4.840.48

a = 10.1Ansm/s²_Ans

A ball is projected at 5i + 20j from a tower of height 5m Find the following: The time of flightThe maximum heightThe maximum horizontal range

Given: u = 20,
s = -5 t = ?,
a = -10
We know,
s = ut + 1/2 at²
-5= 20t - 1/2 x 10 x t²
-5 = 20t - 5t²
5t² - 20t - 5 = 0
t = 4.2s or t = -0.2s
Time of flight = 4.2s
Given: u = 20,
a = -10,
v = 0,
s = ?
We have,
v² = u² + 2as
0 = 400 - 20s
s = 20m
Maximum height = 5 + 20 = 25m
Given: u = 5,
a = 0,
t = 4.2
s = 5 x 4.2sec
s = 21m

State's all three Newton's laws of motion .

1. Newton's First Law of Motion (Law of Inertia):

- "An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an external force."

- In simpler terms, this means that an object will maintain its state of motion (whether at rest or moving) unless an external force acts upon it. If no force is applied, the object will continue its motion or remain at rest.

2. Newton's Second Law of Motion (Law of...

ask mattrab