explain part a An object is initially at x = 0 and moves along the x axis accord
ID: 1630285 • Letter: E
Question
explain part a
An object is initially at x = 0 and moves along the x axis according to the velocity-time graph in Figure. (a) What is the object's acceleration at 15.0 s? a = 0 - 18/5 = -3.6m/s^2 (b) Indicate every time interval(s) for which the speed is decreasing. 4-6 seconds and 13-18 seconds (c) At what time is the object closest to x = 0? The total cumulative area is approximately zero around 11.33 seconds. A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a body of water as shown in Figure below.Explanation / Answer
Acceleration is given by:
a = dV/dt
a = (Vf - Vi)/(tf - ti)
In the graph velocity is constant from t = 9 sec to t = 13 sec, after that velocity start decreasing with constant acceleration,
Now from t = 13 to t = 18 sec velocity decreases with contant rate from V = 18 m/sec to V = 0 m/sec, So
a = (0 - 18)/(18 - 13)
a = -18/5 = -3.6 m/sec^2
Let me know if you have any doubt.
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