Example on Constant Acceleration||Study OF Graph with Constant Acceleration x-t||v-t||a-t

V-t Graph|If A Body Has Constant Acceleration |A-t Graph|Exercise Questions |If Acceleration Of The Body Is Zero |Slope Of Tangent Of X-t Curve- Velocity

Graphs are very useful to represent a physical situation. Various quantities can be easily represented on graphs and other quantites can be determined from the graph. For example the slope of velocity-time graph represents instantaneous acceleration. For a motion with constant acceleration slope of velocity-time graph is constant. If acceleration is changing with time, slope with change and thus velocity-time graph will be a non-linear curve. Further the area of velocity-time graph gives displacement. For the above situation let t_(1) be the time of accelerated motion and t_(2) be the time or retarded motion, then the correct relation is

Graph|V - t Graph|a - t Graph|If Acceleration Of The Body Is Zero|If A Body Has Constant Acceleration|Question

Graph|V - t Graph|a - t Graph|If Acceleration Of The Body Is Zero|If A Body Has Constant Acceleration|Question

Starting from rest a particle is first accelerated for time t_1 with constant acceleration a_1 and then stops in time t_2 with constant retardation a_2. Let v_1 be the average velocity in this case and s_1 the total displacement. In the second case it is accelerating for the same time t_1 with constant acceleration 2a_1 and come to rest with constant retardation a_2 in time t_3. If v_2 is the average velocity in this case and s_2 the total displacement, then

Assertion : In projectile motion at any two positions (v_2-v_1)/(t_2 -t_1) always remains constant. Reason : The given quantity is average acceleration, which should remain constant as acceleration is constant.

A particle moves in a circle of radius 2 cm at a speed given by v =4t, where v is in cms ^-1 and t in seconds. The tangential acceleration at t =1s and total acceleration at t =1s are respiccitively .

Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane: (1) x=-3t^(2)+4t-2andy=6t^(2)-4t (2) x=-3t^(3)-4t andy=-5t^(2)+6 (3) vecr=2t^(2)hati-(4t+3)hatj (4) vecr=(4t^(3)-2t)hati+hatj Are the x and y acceleration components constant? Is acceleration veca constant?

A particle is moving in circular path with constant acceleration. In time t after the beginning of motion the direction of net acceleration is at 30^(@) to the radius vector at that instant. The angular acceleration of the particle at that time t is