JEE Advanced (Single Correct MCQs): Motion - JEE MCQ

Time taken to cross the river t = d/v_5cosθ

_{NOTE :  For time to be minimum, cosθ = maximum⇒θ= 0°}

_{The swimmer should swim due north.}

Shortest route corresponds to vrb perpendicular to river flow

| Average velocity | =

(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.
Further,

Change in velocity = area under the graph

Since, initial velocity is zero, final velocity is 55 m/s.

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).

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.

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.

Average acceleration

To find the resultant of and  , we draw a diagram 

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.

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.