A cart started at t=0, its acceleration varies with time as shown in figure. Find the distance travelled in 30 seconds and draw the position-time graph
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Answer:
It is clear from the graph that the cart has uniform acceleration of 5 ft/s² for initial 10 s then uniform velocity for next 10 s (because acceleration is zero) and then retardation in further next 10 s. Area between graph and time axis gives the change in velocity in the taken time interval Let us consider three parts of each 10 s. In the first 10 s change of velocity = 5 ft/s² x 10 s = 50 ft/s, Since the cart starts from rest at t=0, average velocity in this period = (50+0)/2 =25 ft/s, Distance travelled in this 10 s =25 ft/s x 10 s =250 ft. From t=10 s to 20 s area under the graph is zero, so change in velocity is zero, it means that the velocity at t= 10 s (50 ft/s) remains unchanged. So distance travelled in this period = 50 ft/s x 10 s = 500 ft. From t=20 s to 30 s area under the graph = 5 ft/s² x 10 s = 50 ft/s but it is below the time axis, So this change of velocity is negative. It means that the cart moving with a velocity of 50 ft/s at t= 20 s comes to rest at t=30 s. So average velocity in this period = 25 ft/s and distance travelled = 25 ft/s x 10 s = 250 ft. So the distance travelled = 250 ft + 500 ft+ 250 ft = 1000 ft. The position-time graph may be drawn as follows,
