Kinematics question on a motorcycle

[imy786]

i didnt calculate previos parts..so i just made assumption and put any random figure to carry on the next part of the quesiton

 

[cristo]

i didnt calculate previos parts..so i just made assumption and put any random figure to carry on the next part of the quesiton

Why? You are given sufficient information to answer each part of the question, so there is no need to enter random figures!

 

[hage567]

i didnt calculate previos parts..so i just made assumption and put any random figure to carry on the next part of the quesiton

Why would you do that? You will have to solve all of it eventually. If you do it in the right order it makes everything easier. It will also be less confusing if you don't jump around from one thing to another.

 

[imy786]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------------------------------
2 equations- horizontal and verical motion-
horizontal speed remains constant as there is no acceleration
horizontal speed is= 20cos 11.53= 19.60m/s

vertical motion to calculate vertical speed:

v=
u= 20sin11.52= 4m/s
a= -10
t=2 (time from part c calculated solving quadratc equaiton)

v=u+at
= 4- (20)= -16 m/s = vertical speed
?/

is this vertical speed correct ?

 

[imy786]

need help here...

 

[imy786]

please someone help me do this question URGENT...

 

[hage567]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------------------------------
2 equations- horizontal and verical motion-
horizontal speed remains constant as there is no acceleration
horizontal speed is= 20cos 11.53= 19.60m/s

vertical motion to calculate vertical speed:

v=
u= 20sin11.52= 4m/s
a= -10
t=2 (time from part c calculated solving quadratc equaiton)

v=u+at
= 4- (20)= -16 m/s = vertical speed
?/

is this vertical speed correct ?

Yeah, that seems OK for the vertical velocity. So put that together with the horizontal velocity to get the total velocity. Use the Pythagorean theorem. Please look it up if you don't know how to do it, and give it a try.

please someone help me do this question URGENT...

Then please work everything out to your final answers for your next post so it can be checked all at once. This question seems to have been considered "urgent" for quite some time.

 

[imy786]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------

total velocity= adding both vector velocity

horizontal=19.60 m/s
vertical speed= -16
inverse tan (-16/19.6) = angle= -39.22

total velocity= horizontal +vertical = 19.6- 16 =3.6 m/s

magnitude of velocity is 3.6 m/s at angle -39.2 degrees direction.


is this correct?
angle being a minus ?should that be ok...?

 

[hage567]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------

total velocity= adding both vector velocity

horizontal=19.60 m/s
vertical speed= -16
inverse tan (-16/19.6) = angle= -39.22

total velocity= horizontal +vertical = 19.6- 16 =3.6 m/s

magnitude of velocity is 3.6 m/s at angle -39.2 degrees direction.


is this correct?
angle being a minus ?should that be ok...?

A negative angle is fine, think about which direction the velocity is with respect to the horizontal at the end of the jump.

NO the total velocity is not 3.6 m/s. I told you to use the Pythagorean theorem, do you not know what that is? How do you find the hypontenuse of a right angle triangle? That is really what you are doing here.

 

[imy786]

ow...i was calculating using vectors...
a2+b2= c2...

hage- iv done a leve maths and physics ...so i aint a total beginer...

sqaure root of (19.6^2 + 16^2) = 25m/s is the total velocity at -39.2 degrees
with respect to the horizontal

 

[hage567]

ow...i was calculating using vectors...
a2+b2= c2...

hage- iv done a leve maths and physics ...so i aint a total beginer...

sqaure root of (19.6^2 + 16^2) = 25m/s is the total velocity at -39.2 degrees
with respect to the horizontal

Well when I say use Pythagorus and you don't, what else am I to assume!?

That answer looks reasonable to me.

 

[imy786]

you have been so helpful...