Question below?
1 Answer
(a)
(b)
(c)
(d)
Explanation:
According to the given image, we're asked to find
-
(a) the velocity of the object from
#t = 0# to#t = 2# #"s"# -
(b) the velocity of the object from
#t = 8# #"s"# to#t = 12# #"s"# -
(c) the velocity of the object from
#t = 5# #"s"# to#t = 12# #"s"# -
(d) what appears to be the average velocity for the time intervals
(i)
#t in [0, 4color(white)(l)"s"]# (ii)
#t in [0, 6color(white)(l)"s"]# (iii)
#t in [0, 9color(white)(l)"s"]#
(a)
The velocity of the object from the time interval
#t in [0, 2color(white)(l)"s"]# I'll assume each hash mark on the displacement axis is
#5# #"m"# .The velocity
#v# for a time interval#t# is
#v = (x-x_0)/t# where
#x# is the final position and#x_0# is the initial position.On the interval from zero to two seconds, the initial position (at
#t = 0# ) appears to be#25# #"m"# , and the final position (at#t = 2# #"s"# ) appears to be#10# #"m"# , so we have
#v = (10color(white)(l)"m" - 25color(white)(l)"m")/(2color(white)(l)"s") = color(red)(-7.5# #color(red)("m/s"#
(b)
The velocity of the object from
#t in [8color(white)(l)"s", 12color(white)(l)"s"]# Again using the equation
#v = (x-x_0)/t# The initial position (
#t = 8# #"s"# ) appears to be#20# #"m"# , and the final position (#t = 12# #"s"# ) appears to be#25# #"m"# (which was its starting position), so we have
#v = (25color(white)(l)"m" - 20color(white)(l)"m")/(12color(white)(l)"s" - 8color(white)(l)"s") = color(blue)(1.25# #color(blue)("m/s"#
(c)
The velocity of the object in
#t in [5color(white)(l)"s", 12color(white)(l)"s"]# The initial position (
#t = 5# #"s"# ) appears to be#10# #"m"# , and the final position (#t = 12# #"s"# ) is#25# #"m"# , as found in the last problem.Thus, we have
#v = (25color(white)(l)"m" - 10color(white)(l)"m")/(12color(white)(l)"s" - 5color(white)(l)"s") = color(red)(2.14# #color(red)("m/s"#
(d)
- (i)
The average velocity on
#t in [0, 4color(white)(l)"s"]# At time
#t = 4# #"s"# , the position is not changing from#t = 2# #"s"# , so it is#10# #"m"# . The initial position (#t = 0# ) is#25# #"m"# , so we have
#v = (10color(white)(l)"m" - 25color(white)(l)"m")/(4color(white)(l)"s") = color(blue)(-3.75# #color(blue)("m/s"#
- (ii)
The average velocity from
#t in [0, 6color(white)(l)"s"]# At
#t = 6# #"s"# , there isn't a discrete line saying what the position is...but we can find it!The position at
#t = 6# #"s"# lies along the velocity line during#t in [5color(white)(l)"s", 8color(white)(l)"s"]# , which is
#v = (20color(white)(l)"m" - 10color(white)(l)"m")/(8color(white)(l)"s" - 5color(white)(l)"s") = color(blue)(3.33# #color(blue)("m/s"# Six seconds minus five seconds is
#color(green)(1# #color(green)("s"# , so this will; be#t# in the equation. We can rearrange the equation to solve for position,#x# :
#x = x_0 + vt# The initial position is
#10# #"m"# , so we have
#x = 10# #"m"# #+ (color(blue)(3.33color(white)(l)"m/s"))(color(green)(1color(white)(l)"s")) = color(purple)(13.3# #color(purple)("m"# Thus, the average velocity on the interval
#t in [0, 6color(white)(l)"s"]# is
#v = (color(purple)(13.3color(white)(l)"m") - 25color(white)(l)"m")/(6color(white)(l)"s") = color(red)(-1.94# #color(red)("m/s"#
- (iii)
The average velocity on the interval
#t in [0, 9color(white)(l)"s"]# The position at
#t = 9# #"s"# is part of the velocity line for the interval#t in [8color(white)(l)"s", 12color(white)(l)"s"]# , which we found to be#1.25# #"m/s"# .Nine seconds minus eight seconds equals
#1# #"s"# , so this is#t# in the equation, so we have
#x = x_0 + vt = 20color(white)(l)"m" + (1.25color(white)(l)"m/s")(1color(white)(l)"s") = color(red)(21.25# #color(red)("m"# The average velocity is thus
#v = (color(red)(21.25color(white)(l)"m") - 25color(white)(l)"m")/(9color(white)(l)"s") = color(blue)(-0.417# #color(blue)("m/s"#
Hope this helped:) (And I also hope it wasn't too confusing...I notice the first three answers are given to you underneath the problem)
