The graphs x(t) below represent the motion of two ladybugs, Kyle(K)(solid graph)

Question

The graphs x(t) below represent the motion of two ladybugs, Kyle(K)(solid graph) and Lea (L) (dotted graph) moving along the same straight line. The initial distance between K and L is 4m 5m 9m At t=3 s. the distance between K and L is ? -2 m 2 m 4 m 6 m what is the Instantaneous velocity of K at t = 1.0 s? Zero 0 75 m/s 1.33 m/s 4.0 m/s What is the instantaneous velocity of K at = o 6.0 s? 0.67 m/s 1.33 m/s -1.33 m/s Zero -0.75 m/s 0.07 m/s What is the instantaneous velocity of L at t = 6.0 s? 0 33 m/s -0.33 m/s Zero -3.0 m/s 3.0 m/s 0.5 m/s What is the average velocity of l between t = 0 and t = 10 s? 02 m/s 1.33 m/s 90m/s Zero 0.9m/s 2.0m/s What is the displacement of K between i = 1.0 s and t = 10.0 s? 14 m 8 m -14 m -8 m 4 m -4 m 10 m -10 m What is the distance covered by K between t = 1.0 s and t = 10.0 s? 14 m 8 m -14 m -8 m 4 m -4 m 10 m -10 m How can the position and the velocity of K be described at the moment that l passes through the point x=0? xk 0 xk 0; vk = 0 What Statement correctly describes the sign of the acceleration of K at 1.6 s? a >0 a 0 a0 a

Explanation / Answer

Part 1= 9m

Part 2 =6m

Part 3 =zero

Part 4= The speed of K is -1.33m/s as we see the graph descending we are sure it will be in negative and -0.75m/s cant be the velocity

Part 5=Instantaneous velocity zero

Part 6=Instantaneous velocity zero

Part 7= -4m will be the displacement

Part 8= 10m will be the distance

Part 9= xk<0;vk>0

Part 10,Part 11 and Part 12=accelaration will be 0 because the velocity is increasing

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