Classes are canceled due to snow, so you take advantage of the extra time to con
ID: 1653313 • Letter: C
Question
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled, and take the modified sled to a large, flat, snowy field. You ignite the rocket, and observe that the sled accelerates from rest in the forward direction at a rate of 13.5 m/s^2 for a time period of 2.90 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 4.15 m/s^2. After the rocket turns off, how much time does it take for the sled to come to a stop? By the time the sled finally comes to a rest, how far has it traveled from its starting point?Explanation / Answer
A) Given intial at rest means intial velocity u1=0 m/s
forward acceleration a1 = 13.5m/s2 for time pertiod t1= 2.90 s.
Velocity of sled after 2.90s (when rocket engine abruptly shuts off) will be v1 = u1 + a1t1
v1 = 0+(13.5*2.90) = 39.15 m/s
Distance travelled during 2.90 s will be given by s1 = u1t1 +0.5*a1(t1)2
s1 = 0+0.5*13.5*(2.90)2 = 56.7675 m
At the end sled stops means final velocity zero, let t2 is time taken by sled to stop after rocket turns off, during this acceleration is backward that is a2 = -4.15 m/s2
Use v=u+at
0 = 39.15 + (-4.15*t2)
t2 = 39.15/4.15 = 9.43 s (Answer)
Distance travelled during last 9.43 s will be given by s2 = v1t2 +0.5*a2t22
s2 = (39.15*9.43) +(0.5*-4.15*(9.43)2)
s2 = 184.67 m
B) total distace travelled = s1 +s2 = 56.7675+184.67 = 241.44 m (answer , up to two significant figure)