HC Verma Questions and Solutions: Chapter 3- Rest and Motion: Kinematics- 1 | HC Verma Solutions - JEE PDF Download

70 km/h towards south-west
Final velocity,  Initial velocity,  Change in velocity, 

It is towards southwest, as shown in the figure.

The slope of the x–t graph gives the velocity. In the graph, the slope is constant from t = 0 to t = t₀, so the velocity is constant. After t = t₀, the displacement is zero; i.e., the particle stops. 

As velocity and acceleration are in opposite directions, velocity will become zero after some time (t) and the particle will return.

Because the value of acceleration is not given, we cannot say that the particle will return after/before 10 seconds. 

Along the string, the velocity of each object is the same.

Velocity is uniform in both cases; that is, acceleration is zero.

Velocity is uniform in both cases; that is, acceleration is zero.


∴  Average velocity,  

Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.

Total energy of any particle = 
Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.
At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.

In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.

Because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.
Time of flight = T = 

For the same u range, 

Horizontal range for the projectile,   Information of the initial velocity is not given in the question

If the man swims at any angle east to the north direction, although his relative speed will increase, he will have to travel a larger distance. So, he will take more time.
If the man swims at any angle west to the north direction, his relative speed will decrease. So, he will take more time. 

(a) the displacement is zero
(d) the average velocity is zero
Displacement is zero because the initial and final positions are the same.
Average velocity = Displacement/Total time = 0
Distance covered = 2πr ≠ 0
Average speed = Distance travelled/Total time taken ≠ 0

(c) the acceleration of the particle is 2a
(d) at t = 2 s, the particle is at the origin
Initial velocity = 

Acceleration = 
At t = 2 s,
x = u(2 s − 2 s) + a (2 s − 2 s)² = 0 (origin) 

(a) Average speed of a particle in a given time is never less than the magnitude of the average velocity.
(b) It is possible to have a situation in which 
(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval, for example, the motion of a particle on a circular track with a constant speed.
Average velocity = Displacement/ Total time
Displacement ≤ Distance
∴ Average velocity ≤ Average speed  In uniform circular motion, speed is constant but velocity is not.
i.e.,  which proves case (b) 
(d) In one complete circle of uniform motion, average velocity is zero. Instantaneous velocity is never zero in the interval.

(b) varying velocity without having varying speed
(d) nonzero acceleration without having varying speed
Velocity and acceleration are vector quantities that can be changed by changing direction only (keeping magnitude constant).

(a) If the velocity and acceleration have opposite signs, the object slows down.
(b) If the position and velocity have opposite signs, the particle moves towards the origin.
(d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.
(a) Acceleration is given by

(b) If the position and velocity have opposite signs, the particle moves towards the origin. It can be explained by following figure:
 
(c) If the velocity is zero in a certain time interval, then the change in the velocity in that time interval will also be zero. As acceleration is rate of change of velocity, it will also be zero at an instant in that time interval.

(b) The acceleration at t = 0 may be zero.
(c) If the acceleration is zero from t = 0 s to t = 10 s, the speed is also zero in this interval.
(d) If the speed is zero from t = 0 s to t = 10 s, the acceleration is also zero in this interval.
(b) Acceleration will be zero only when the change in velocity is zero.
(c) Since the acceleration is zero from t = 0 s to t = 10 s, change in velocity is 0.
Velocity in this interval = Initial velocity = 0
Also,
Speed in this interval = Initial speed = 0
(d) From t = 0 s to t = 10 s, speed is zero.
Here, velocity is zero and initial velocity is zero.
So, the change in velocity is zero; i.e., acceleration is zero. 

The magnitude of the velocity of a particle is equal to its speed.
(a) Velocity being a vector quantity has magnitude as well as direction, and magnitude of velocity is called speed.
(b) Average velocity = Total displacement/Total time taken 
Average speed = Total distance travelled/Total time taken 
Distance ≥ Displacement
∴ Average speed ≥  Average velocity
The magnitude of average velocity in an interval is not always equal to its average speed in that interval. 
(c) If speed is always zero, then the distance travelled is always zero. Hence, the total distance travelled and the average speed will be zero.
(d) If the speed of a particle is never zero, the distance travelled by the particle is never zero. Hence, the average speed will not be zero.

(a) The particle has a constant acceleration.
(d) The average speed in the interval 0 s to 10 s is the same as the average speed in the interval 10 s to 20 s.
Explanation:
(a) The slope of the v–t graph gives the acceleration. For the given graph, the slope is constant. So, acceleration is constant.
(b) From 0 to 10 seconds, velocity is in positive direction and then in negative direction. This means that the particle turns around at t = 10 s.
(c) Area in the v–t curve gives the distance travelled by the particle.
Distance travelled in positive direction ≠ Distance travelled in negative direction
∴ Displacement ≠ Zero
(d) The area of the v–t graph from t = 0 s to t = 10 s is the same as that from t = 10 s to t= 20 s. So, the distance covered is the same. Hence, the average speed is the same.

The particle has come to rest 6 times.
Explanation:
(a) The slope of the x–t graph gives the velocity. Here, 6 times the slope is zero. So, the particle has come to rest 6 times.
(b) As the slope is not maximum at t = 6 s, the maximum speed is not at t = 6 s.
(c) As the slope is not positive from t = 0 s to t = 6s, the velocity does not remain positive.
(d) Average velocity =  Total displacement/Total time taken 

For the shown time (t = 6 s), the displacement of the particle is positive. Therefore, the average velocity is positive.

The acceleration of S₂ with respect to S₁ may be anything between zero and 8 m/s².
Explanation: