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1. So I began working the question (in attachment). When I was working out the time, I assumed the initial velocity was zero but the next question proves that it is not. Please help because I am really confused.
Thanks
Attached Files
2. physics motion.docx (23.1 KB, 76 views)
3. The initial vertical velocity is zero. The initial horizontal velocity isn't. If you carefully split it into vertical and horizontal problems, it should come out OK.
4. initial velocity in the y-component is indeed zero but its constant in the x-component. knowing the time it took and the horizontal distance you should be able to work out the initial velocity
5. ok so i have
a- 9.8
s- 3.0
v- 0
So could i use the equation v²=u²+2as to find the initial velocity.
6. horizontal:
v=
u=
a=0
t=
s=3

vertical:
v=
u=0
a=9.8
t=
s=0.2

s=ut+0.5at2

0.2=0+0.5*9.8*t2

t2=0.2/4.9
t2=0.0408
t=0.202 secs

initial vertical velocity = 0
initial horizontal velocity = u

s=ut+0.5at2

3=0.202u+0.5*0*0.2022
u=3/0.202
u= 14.8 m/s

t2 means t squared, my keyboard is broken
7. thanks
(Original post by Artemis 97)
horizontal:
v=
u=
a=0
t=
s=3

vertical:
v=
u=0
a=9.8
t=
s=0.2

s=ut+0.5at2

0.2=0+0.5*9.8*t2

t2=0.2/4.9
t2=0.0408
t=0.202 secs

initial vertical velocity = 0
initial horizontal velocity = u

s=ut+0.5at2

3=0.202u+0.5*0*0.2022
u=3/0.202
u= 14.8 m/s

t2 means t squared, my keyboard is broken
8. Sorry to be a pain but, i have just come across another question and it is really confusing me because I'm not sure how to work out the time if the cars are moving.
Attached Files
9. physics motion 2.docx (16.2 KB, 711 views)
10. (Original post by naandw94)
Sorry to be a pain but, i have just come across another question and it is really confusing me because I'm not sure how to work out the time if the cars are moving.
Can you find relative speed of the vehicles?
11. (Original post by jaroc)
Can you find relative speed of the vehicles?
would that not just not be 30ms^1 for the car and 20ms^1 for the lorry so instead I find the time using time=distance/speed?
12. Car_______________Lorry
v=30______________v= 20
s=x_______________s= x-95
t=t________________t= t

v=s/t
30=x/t
20= (x-95)/t
x=30t
x=20t + 95
30t=20t+95
10t=95
t=9.5s

30=x/t
t=9.5
30=x/9.5
x= 285

car travels 285m
lorry travels 190m
13. (Original post by naandw94)
would that not just not be 30ms^1 for the car and 20ms^1 for the lorry
Relative speed simply means the difference in their velocities.

I find the time using time=distance/speed?
All right, you can approach the question without the concept of relative speed.

In the time taken by the car to draw level with the lorry, the lorry will have travelled some distance and the car will travel a distance greater by 100m. Can you see that?
14. just set up a set of equations.

the lorry has traveled a distance 95+vt by the time the car travels the necessary distance to catch up which is vt. so like what has been mentioned before, while the car is traveling to level with the lorry, the lorry is still moving at a speed 20m/s, i.e. its not like it stops while the car is moving. as the both cars are going at their respective speeds, they both travel distances vt, which as you can tell depends on their speeds. since time t is the same you get a very nice algebraic expression
95+20t=30t

the b) part is just a simple v=d/t calculation

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