Kinematics Test - 11

Step 1:  Calculate velocity of car P and Q

Step 2:  Time at which both have the same velocity

Let t₀​ be the time when V_P​ and V_Q​ are same 

Equate equation (1) and (2)

Because the slope is highest at C, v = dt/ds​ is maximum.

Area under the graph represents the change in velocity so

t = 2 α x² i.e. x² = (t / 2α) i.e. x² = constant × t 2α is constant.

This equation is same as equation of parabola y² = 4 ax hence shape of graph is of parabola.

Displacement is the shortest distance travelled from the initial point to the final point. After half revolution the displacement is the length of the diameter which is equal to 2r. Hence the displacement is 2r.

The vertical axis will represent the acceleration of the object. The slope of the acceleration graph will represent a quantity called the jerk. This jerk is the rate of change of the acceleration. The area under this acceleration graph represents the change in velocity. Also, this area under the acceleration-time graph for some time interval will be the change in velocity during that time interval. Multiplying this acceleration by the time interval will be equivalent to finding the area under the curve.

Resultant of two vectors is given by parallelogram law of vector addition. As per this law, the resultant vector is R = √(​A² + B² + 2ABcosθ). Here A and B are the two vectors and θ is the angle between them. You can see that R will be maximum when θ is equal to zero that is when the 2 vectors are parallel to each other.

average velocity = {(displacement) / (time)} = {x / (T/6)} = (6/T) x   (1) 

in time T, particle completes 360° hence in (T/6) time 60° will be covered 

θ = (x/R) hence (π/3) = (x/R) i.e. x = 1.04 R 

from (1), average velocity = (6/T) (1.04R) = (6R/T)

If the displacement is along a circular path, then the direction of linear velocity at any instant is along the tangent at that point.
therefore, the linear velocity will be ω.r