A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction
Time taken to cross the river t = d/v5cosθ
NOTE : For time to be minimum, cosθ = maximum⇒θ= 0°
The swimmer should swim due north.
A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/hr is
Shortest route corresponds to vrb perpendicular to river flow
In 1.0 s, a particle goes from point A to point B, moving in a semicircle of radius 1.0 m (see Figure). The magnitude of the average velocity
| Average velocity | =
A ball is dropped vertically from a height d above the ground.
It hits the ground and bounces up vertically to a height d/2.
Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the ground as
(a) KEY CONCEPT Before hitting the ground, the velocity v is given by v2 = 2 gd (quadratic equation and hence parabolic path) Downwards direction means negative velocity. After collision, the direction become positive and velocity decreases.
A particle starts sliding down a frictionless inclined plane.
If Sn is the distance travelled by it from time t = n – 1 sec to t = n sec, the rati o Sn/Sn+1 is
A body starts from rest at time t = 0, the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be
Change in velocity = area under the graph
Since, initial velocity is zero, final velocity is 55 m/s.
The velocity-displacement graph of a particle moving along a straight line is shown
The most suitable acceleration-displacement graph will be
The equation for the given v-x graph is
On comparing the equation (ii) with equation of a straight line
y = mx + c
i.e., they-intercept is negative The above conditions are satisfied in graph (a).
Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed w. The discs are in the same horizontal plane. At time t = 0, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is vr. In one time period (T) of rotation of the discs, vr as a function of time is best represented by
At t = 0, the relative velocity will be zero.
At , the relative velocity will be maximum in magnitude.
At , the relative velocity will be zero.
At , the relative velocity will be maximum in magnitude
At t = T, the relative velocity again becomes zero.
Consider a disc rotating in the horizontal plane with a constant angular speed w about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/8 rotation, (ii) their range is less than half the disc radius, and (iii) w remains constant throughout. Then .
The x-coordinate of
This horizontal distance travelled will be greater than any point on the disc between O and P. Therefore the landing will be in unshaded area. In the same way, the horizontal distance travelled by Q is always less than that of any point between O and R. Therefore the landing will be in unshaded area.
A particle is moving eastwards with a velocity of 5 m/s. In 10s the velocity changes to 5 m/s northwards. The average acceleration in this time is
To find the resultant of and , we draw a diagram
A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. NO other information is available about its motion at intermediate times (0 < t< 1). If a denotes the instantaneous acceleration of the particle, then:
Note : a cannot remain positive for all t in the interval 0 < t < 1. This is because since the body starts from rest, it will first accelerate. Finally it stops therefore a will become negative. Therefore a will change its direction. Options (a) and (d) are correct.
Let the particle accelerate uniformly till half the distance (A to B) and then retard uniformly in the remaining half distance (B to C).
The total time is 1 sec. Therefore the time taken from A to B is 0.5 sec.
For A to B
Note : Now, if the particle accelerates till B2 then for covering the same total distance in same time, acceleration should be less than 4 m/s2 but |deceleration| should be greater than 4 m/s2. And if the particle accelerates till B1, then for covering the same total distance in the same time, the acceleration should be greater than 4 m/s2 and | deceleration | < 4 m/s2.
The same is depicted by the graph.
So, the | acceleration | must be greater than or equal to 4 m/s2 at some point or points in the path.
The coordinates of a particle moving in a plane are given by x(t) = a cos (pt) and y (t) = b sin (pt) where a, b (< a) and p are positive constants of appropriate dimensions. Then
Squaring and adding (1) and (2), we get,
∴ The path of the particle is an ellipse.
From the given equations we can find,
ax and vy become zero Only
vx and ay are left, or we can say that velocity is along negative x-axis and acceleration along negative y-axis.
Hence, at velocity and acceleration of theparticle are normal to each other.
At t = t , position of the particle
Therefore, acceleration of the particle is always directed towards origin.
(0, b). Therefore, the distance covered is one fourth of the elliptical path and not a.