xim) (C) 6s (D) 12s (E) 25s 16 10 t(s) The graph above represents position x ver

Question

xim) (C) 6s (D) 12s (E) 25s 16 10 t(s) The graph above represents position x versus time t for an object being acted on by a constant force. The average speed during the interval between I s and 2s A rock of mass m is thrown horizontally off a building from a height h as shown above. The speed of the rock as it leaves the thrower's hand at the edge of the building is I. is most nearly (A) 2 m/s (B) 4 m/s (C) 5 m/s (D) 6 m/s (E) 8 m/s How much time dees it take the rock to travel the edge of the building to the ground? from (A) Vhv (B) h va (C) hv/g (D) 2h/g Two objects, X and Y, accelerate from rest with the same constant acceleration. Object X accelerates for twice the time as object Y. Which of the following is true of these objects at the end of their respective S. Time periods of acceleration? object Y object Y object Y 2. The motion of a particle along a straight line is (A) Objedt X is moving at the same speed as (BObjea X is moving four times faster than (C Giejoot X has traveled the same distance as represented by the position versus time graph above. At which of the labeled points on the graph is the magnitude of the accelcration of the particle greatest? (A) A (B) B (D) D (E) E (D) Objoct X has traveled twice as far as object Y (E) Object X has traveled four times as far as object Y An object is thrown with a horizontal velocity of 20 m/'s from a cliff that is 125 m above level ground. air resistance is negligible, the time that it takes the object to fall to the ground from the cliff is most nearly 3. (A) 3s (B) 5s

Explanation / Answer

1. form the graph,

between 1 and 2 seconds

displacement = 8 -2 = 6 m

time taklen = 1 s

hence average speed = 6 m/s

Option D) 6 m/s

2. Acceleration is double derivative of p[osition time graph]

now at positions A and B, the graph is linear so velocity is constant and acceleration is 0

same is the case with positions D and E

hence

maximum acceleration is at C

OptionC) C

3. as the object is thrown with horizontal velocity its initial vertical velocity is 0

hence time taken to fall = t

height of cliff, h = 125 m

hence

h = 0.5gt^2

125*2/g = t^2

t = 5.048 s

so, answer is option B) 5 s

4. similairly to the last problem

for a cliff height h and no vertical vcelocity intially

time takne to fall t = sqroot(2h/g)

hence option E) sqroot(2h/g)

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