In problem 35 the textbook solution is wrong for the way the problem is stated.
This was not an assigned HW problem (but the diligent among you may have tried
it and gotten confused). Here is what is wrong.
The problem as stated:
A train is travelling south at 24.0 m/s when the brakes are applied. It slows
down with a constant acceleration to a speed of 6.00 m/s in a time of 9 s.
(a) Draw a graph of vx versus t for a 12 s time interval (starting 2 seconds
before the brakes are applied and ending 1 s after the brakes are released).
Let the x-axis point to the north.
The textbook answer for this part in the back
of the book is shown to the right. This would
have been correct had the problem instead said:
Let the x-axis point to the south.
As stated, however, with the x-axis pointing north and the train traveling south the
two velocities both also point south, so they should be vix = –24 m/s and
vfx = –6 m/s (the speeds given in the statement of the problem would still be
positive). With the velocities negative the solution for part (a) should be as shown
here.
0
For part (b) which asks for the acceleration
=
ax
=
ax
v fx − v ix
=
∆t
−6
−6
m
m
− ( −24 )
s
s
9s
m
m
m
+ 24
18
m
s
s
s
= = 2 2
9s
9s
s
–8
–16
–24
–
Which is also the slope of the vx versus t plot during the 9 s (note that positive).
The textbook solution is right for the case drawn here (positive slope).
For part (c) the displacement during the 9 s acceleration phase is the shaded “area”
between the curve and the axis. This will be
1
m
m
∆x = ( −18 )(9 s) + ( −6 )(9 s) =−135 m
2
s
s
0
Or algebraically since,
1
∆x= v ix ∆t + a x ∆t 2
2
m
1 m
∆x =−
( 24 )(9 s) + (2 2 )(9 s)2 =−135 m
s
2 s
–8
–16
–24
–
In either case the distance travelled is the magnitude of this or 135 m.