Bob starts off at rest at x=0. Then, Bob increases his speed at a constant rate,
ID: 2146762 • Letter: B
Question
Bob starts off at rest at x=0. Then, Bob increases his speed at a constant rate, so that after 4 s, he is traveling8m/s in the +x direction. Bob then decreases his speed at a constant rate for 2 s, ending up back at rest.
a.)Draw Bob?s vxvs. t graph, and use it to answer the following:
? at t=0 s, vx=m/s
? from 0 s to 4 s, this graph is most nearly shaped like:A B C D E F G
? at t=4 s, vx=m/s
? from 4 s to 6 s, this graph is most nearly shaped like:A B C D E F G
? at t=6 s, vx=m/s
b.)Draw Bob?s axvs. t graph, and use it to answer the following:
? from 0 s to 4 s, this graph is most nearly shaped like:A B C D E F G
? and ax=m/s2
? from 4 s to 6 s, this graph is most nearly shaped like:A B C D E F G
? and ax=m/s2
c.)Find Bob?s position at t = 4 s and 6 s, using graphical methods.
d.)Repeat part c, using the constant acceleration eqns.(This is a good doublecheck that you did part c right.)
e.)Draw Bob's x vs. t graph.
Use your results from parts c-e to answer the following:
? at t=0 s, x =m
? from 0 s to 4 s, this graph is most nearly shaped like:A B C D E F G
? at t=4 s, x =m
? from 4 s to 6 s, this graph is most nearly shaped like:A B C D E F G
? at t=6 s, x =m
Explanation / Answer
You have posted a very long ques.,a question with so many parts.However i am giving you solution and concept which you can use to find all the parts
u(initial velocity)=0
v=u +a*t
For 4 sec,
v(final velocity)=8 m/s
t(time)=4 s
8=0+4*a
8=4a
a=2m/s^2(CONSTANT,FROM 0 TO 4 SEC)
Rest 2 sec,
u=8m/s
v=0(as he comes to rest)
t=2 sec
v=u+a*t
0=8-2*a
a=-4m/s^2 (CONSTANT,FROM 4 TO 6 SEC)
From 0 to 4 sec,the velocity graph is B and acceleration graph is A
From 4 to 6 sec,the velocity graph is c and acceleration graph is A
Use s=ut+0.5*a*t^2 to calculate position at any point.