Question

Observe the graphs of the motion. Consider only the portion of the graph from just after the cart was released to just before it was stopped. Find the position on each graph where the cart turned around and started back down. What is the value of the velocity at that point? What is the value of the acceleration at that point? Is there any significant difference between this acceleration graph and the acceleration graphs from steps 15 and 18? Why or why not? 21.

Answer 1

When the cart turned around and started back, the displacement of the cart will decrease with time as it will move towards its starting point. And from this point velocity will also get reversed.

#From the graph , we find that the at t=2.1s , the cart turned around back.

900,0 780) Time (s)

# At this point the velocity of the cart is Zero.

900,0 780) Time (s)

#slope of the velocity -time graph is m=-0.5537 m/s².

So at that point the acceleration of the cart is a= -0.5537 m/s².

# Since the velocity -time graph is straight fit, the slope is constant and hence acceleration will also constant. so there is no difference between the steps 15 & 18.