question_answer

An object has uniformly accelerated motion. The object always slows down before the time, when its velocity becomes zero. Prove this statement graphically, when (a) both u and a are positive (b) \[u=-ve\] and \[a=+ve\](c) \[u=+ve\]and \[a=-ve\] and (d) both u and a are negative.

Answer:

                (a) When both \[u\] and a are + ve : In such a case, the \[\upsilon -t\]graph will be as shown in Fig. (a). At the time corresponding to point A, the velocity becomes zero. It can be seen that before this time, the velocity is negative but its magnitude decreases with time till it becomes zero at A. (b) When \[u\] is - ve and a is + ve: In this case, graph will be shown in Fig. (b). At the time corresponding to point A, the velocity becomes zero. It can be seen that before this time the velocity is - ve but its magnitude decreases with time till it becomes zero at A. (c) When \[u\] + ve and a is -ve: In such a case, graph between \[\upsilon \]and t will be as shown in Fig. (c). Again at A, velocity is zero. The velocity decreases before the time corresponding to point A.     (d) When both \[u\] and a are - ve: In this case, \[\upsilon -t\]graph will be as shown in Fig. (d). If we produce graph backwards, it meets the time-axis at point A. Before this time, velocity is +ve and decreases till it becomes zero at point A.