I need help with these two physics problems, kinematic 1) A rocket is attached t

Question

I need help with these two physics problems, kinematic

1) A rocket is attached to a toy car that is confined to move in the z-direction. At time to 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +z-direction at 4.25 m/s2. At t1 = 3.60 s, the rocket fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 due to friction. la. (5 points) How many seconds elapse from the time the rocket is lit to the time the car comes to rest? lb. (5 points) What is the car's displacement during its entire trip? lc. (5 points) What is the toy car's average velocity over its entire trip? ld. (5 points) What is the toy car's average acceleration over its entire trip? le. (5 points) Assuming that the car starts at the origin (z = 0 m), sketch a position (m) versus time (s) graph of the car's trip from start to finish. Be sure to label both axes and scale them appropriately. Label at least 3 coordinate pairs that fall on the curve. If. (5 points) Sketch a velocity (m/s) versus time (s) graph of the car's trip from start to finish. Be sure to label both axes and scale them appropriately. Label at least 3 coordinate pairs that fall on the curve. 2) A small cannonball is shot straight up from the ground with a launch speed of 26.3 m/s. 2.00 seconds after the cannonball is launched, a ping pong ball is dropped from a height of 36.5 m. Both objects only move in the y-direction 2a. (10 points) Where (with respect to the ground) do the cannonball and the ping pong ball pass one another? 2b. (5 points) How fast and in what direction are each of the balls traveling when they pass one another? 2c. (5 points) Which ball hits the ground first? How long after the first ball hits the ground does the second ball hit the ground?

Explanation / Answer

1)Given,

t1 = 3.6s ; a1 = 4.25 m/s^2 ; a2 = -1.95 m/s^2

a)from eqn of motion

v = u + at

v1 = 4.25 x 3.6 = 15.3 m/s

when it stops, vf = 0 ; using the same eqn

0 = 15.3 - 1.95 t2 => t2 = 15.3/1.95 = 7.85 s

t = t1 + t2 = 3.6 + 7.85 = 11.45 s

Hence, t = 11.45 s

b)we know that

S = ut + 1/2 at^2

s1 = 0 + 1/2 x 4.25 x 3.6^2 = 27.54 m

s2 = 15.3 x 7.85 - 1/2 x 1.95 x 7.85^2 = 60 m

S = s1 + s2 = 27.54 + 20 = 87.54 m

Hence, S = 87.54 m

c)Cars average velocity is:

Vav = (15.3 + 0)/2 = 7.65 m/s

Vav = 7.65 m/s

d)a(av) = (4.25 - 1.95)/2 = 1.15 m/s^2

a(av) = 1.15 m/s^2

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