The displacement-time graph of a moving particle with constant acceleration is shown in the figure. The velocity-time graph is given by
Correct option is A.
At $$t$$=$$0$$, slope of the $$x-t$$ graph is zero; hence, velocity zero at $$t$$=$$0$$. As time increases, slope increases in negative direction; hence, velocity increases in negative direction. At point '1', slope changes suddenly from negative to positive value; hence, velocity changes suddenly from negative to positive and then velocity starts decreasing and becomes zero at '2'. Option (a) represents all these clearly.
The velocity-time graph AC of a moving particle is shown in the figure below. The acceleration of the particle is:
The acceleration- time graph of a particle moving along a straight line is as shown in figure. At what time the particle acquires its initial velocity?
Velocity (V) versus displacement (S) graph of a particle moving in a straight line is as shown in the figure. Given Vo=4 m/s and So=20 m.
Corresponding acceleration (a) versus displacement (S) graph of the particle would be
The displacement-time graph of a moving particle with constant acceleration is shown in the figure. The velocity-time graph is given by
Velocity versus displacement graph of a particle moving in a straight line is as shown in figure. The acceleration of the particle is:
