Kinematics question on a motorcycle

[imy786]

well if S is not 0, what is it then?
iv put s as -9.8 before that is incroct as well..

need help??

 

[hage567]

s represents the vertical displacement. 9.8 is the acceleration due to gravity. They are not the same thing.
Here's some advice: keep track of the units of all your numbers. You must keep your variables straight. Try to think about the equations you are using and what they mean. Randomly guessing what numbers to put in is only going to cause more confusion.

So, if the height of the ramp above the ground is 20 meters, what is his vertical displacement when he lands on the ground on the other side of the river? That is what you need to put into your equation. See what you get now.

 

[imy786]

-20 is displacemet

is that right

 

[hage567]

Yes! That is right.

 

[imy786]

s= - 20
a= -10
u= 4
t=?
s= ut+1/2 at^2
-20= 4t-5t^2
4t-5t^2+ 20= 0
solving this quadrtic will give value for two time.

 

[hage567]

Yes, but you should get one positive root and one negative root. Which one do you think you should use?

 

[imy786]

(c) Calculate how far from the river the motorcycle lands.
[need help on this part]

i think we have to take horizontal motion into consideration-
s= ?
u= 20 cos 11.53...= 19.6
a= 0 (horizontal no accelration i think)
t= (do i use the t from previos answer (b) ]

then using this equation of motion

s= ut+1/2 at^2

is this correct?

 

[imy786]

urgent help needed...please help me do the quesion

 

[cristo]

(c) Calculate how far from the river the motorcycle lands.
[need help on this part]

i think we have to take horizontal motion into consideration-
s= ?
u= 20 cos 11.53...= 19.6
a= 0 (horizontal no accelration i think)
t= (do i use the t from previos answer (b) ]

then using this equation of motion

s= ut+1/2 at^2

is this correct?

This is the correct method, presuming that the time you calculated in part b is correct.

 

[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.

taking horizontal motion-

v=?
u=
a=o
s=

but if we use v2= u2+ 2as
final velocity would equail intital velocity by calculation (if a=0)

need help?

 

[cristo]

No, the angle will be different. You need to set up two equations for this; one in the horizontal, one in the vertical.

 

[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 motion -

v=
u=?
a=0
s=

what would s and v be for horiznotal motion...

 

[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 motion -

v=
u=?
a=0
s=

what would s and v be for horiznotal motion...

You tell me what you think they should be.

You know the initial horizontal velocity. If there is no acceleration in the horizontal direction, what would it be at the end of the jump?

You can easily find s, because you know the total time of flight (you found that already). So how do you find distance if you know the time and the velocity?

 

[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 motion -

v=
u=?
a=0
s= 10 m (eg assuming this is the value from b)
t= 2 (leg ets assume we calculated t as 2 secs)
if there is no acceratlion then u=V

 

[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 motion -

v=
u=?
a=0
s= 10 m (eg assuming this is the value from b)
t= 2 (leg ets assume we calculated t as 2 secs)
if there is no acceratlion then u=V

Why are we assuming anything? I'm confused as to what you really trying to find. You know what t is from part b. I don't know where the 10 m comes from, I don't see it in any previous post. Isn't s what you were supposed to find for part c? Did you do that? That's the horizontal distance, right?

You don't need s it to do part d anyway. What is the vertical component of velocity at landing? That's what you should be working on, since you already know the horizontal component.

 

[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...