A particle is projected from the origin #(0,0)# in the #xy# plane at angle of #37^o# with the x axis with a speed of #10# #ms^-1#, its acceleration being #(-5 hati + 10 hatj)# #ms^-2#. What is the speed of the particle when the x-coordinate is again zero?
Answer given in the book is #22# #ms^-1# .
Please explain with steps.
Answer given in the book is
Please explain with steps.
4 Answers
Explanation:
We can split this problem up into
We can find the components of the initial velocity using the equation
for
for
We're given that the equations for the components of acceleration are
Using the initial velocity components, and integrating these acceleration equations, we have for the velocity equations
Integrating these equations, the position equations are
We must find the time when
Using this time, we can find the components of the velocity by
The magnitude of the velocity (the speed) is thus
rounded to
Correct me if I made a mistake, but I think this answer should be correct.
See the detailed explanation and working below.
Explanation:
(the units given are a bit odd, in that they mix
So our object is projected from the origin into the first quadrant - top right side of the Cartesian plane, at
This is a different picture from our usual projectile motion problems, because the acceleration is usually provided by gravity and is downward, but it's still just a projectile motion problem, with the exception that there are accelerations in both directions, not just one. We don't know, or need to know, the mass of the particle, because gravity is not involved. We are told its acceleration.
The components of the initial velocity will be given by
We can consider each dimension independently. The position in the x direction will reach a maximum then return to
So the total time will be
The final velocity in the x direction,
The final velocity in the y direction,
The magnitude of the final velocity will be given by:
This is not the answer given in the question, but other answerers get the same, so I assume the textbook is incorrect.
I made my own attempt at this and got
I think the book forgot the
DISCLAIMER: LONG ANSWER!
Notice how if you invert the coordinate axes, i.e. reflect over
GIVEN DATA IN NEW COORDINATES
We will do that, and obtain:
#vecv_i = "10 m/s"#
#veca_x = "10 m/s"^2#
#veca_y = -"5 m/s"^2#
#vecx_i = "0 m"#
#vecy_i = "0 m"#
#theta = 90^@ - 37^@ = 53^@# with respect to (w.r.t) the new#x# axis
Instead of solving for when
DERIVATION OF KINEMATICS EQUATIONS
Integrate the acceleration w.r.t. time:
#vecv_(fx) = int veca_x dt = veca_(x)t + vecv_(ix)# #" "" "bb((1))#
#vecv_(fy) = int veca_y dt = veca_(y)t + vecv_(iy)# #" "" "bb((2))#
Integrate one more time w.r.t time:
#vecx_f = int veca_(x)t + vecv_(ix) dt = 1/2veca_(x)t^2 + vecv_(ix)t# #" "" "bb((3))#
#vecy_f = int veca_(y)t + vecv_(iy) dt = 1/2veca_(y)t^2 + vecv_(iy)t# #" "" "bb((4))# since
#(vecx_i, vecy_i) = (0,0)# .
SOLVING FOR THE TIME OF FLIGHT
Now, using
#1/2veca_(y)t + vecv_(iy) = 0#
#=> t = -(2vecv_(iy))/(veca_y)#
#= -(2cdot"10 m/s" cdot sin(53^@))/(-"5 m/s"^2)#
#=# #"3.195 s"#
VELOCITY COMPONENTS
This means from
#vecv_(fx) = ("10 m/s"^2)("3.195 s") + "10 m/s" cdot cos(53^@)#
#=# #"37.96 m/s"#
#vecv_(fy) = (-"5 m/s"^2)("3.195 s") + "10 m/s" cdot sin(53^@)#
#=# #-"7.986 m/s"#
FINAL VELOCITY
The final velocity I get is:
#color(blue)(vecv_f) = sqrt((37.96)^2 + (-7.986)^2)#
#=# #color(blue)("39 m/s") ne 22#
ERROR IN THE BOOK!
Interestingly enough, if we forgot the
#=> t = -(color(red)(1)vecv_(iy))/(veca_y)#
#= -(color(red)(1)cdot"10 m/s" cdot sin(53^@))/(-"5 m/s"^2)#
#=# #"1.597 s"#
And we would have gotten:
#vecv_(fx) = ("10 m/s"^2)("1.597 s") + "10 m/s" cdot cos(53^@)#
#=# #"21.991 m/s"#
#vecv_(fy) = (-"5 m/s"^2)("1.597 s") + "10 m/s" cdot sin(53^@)#
#=# #-1.38 xx 10^(-31)# #"m/s" ~~ 0#
The final velocity I get would then be:
#color(blue)(vecv_f) ~~ vecv_(fx)#
#=# #color(blue)("22 m/s")#
So yes, I think the book has an error...
The
Movement along
Using the kinematic expression
we get
Using the given condition we get
Ignoring root
Now using the kinematic expression
we get for above value of
Movement along
using (1) we get at the requited
Speed
Hope this helps.