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# m1 kinematics-help plz watch

1. A, B and C are three points on a straight line such that AB=80m, and BC=60m. A car is travelling with uniform acceleration passes A, B and C at times t=0, t=4s and t=6s, respectively. Modellng the car as a particle find it accelerationand its velocity at A
2. em...any help wud b appreciateddddddd.....i b stuck on this Q 4 like an hour now..o.O
3. Use the SUVAT equations.

Assume the car has velocity u at A and acceleration a throughout.

When t = 0, s = 0.

When t = 4, s = 80

When t = 6, s = 140.

Form and solve some simultaneous equations.
4. em..im sorry, but i really dont get that
5. Right, protege know the answers and will look after you.

Do you know the equation ?
6. (Original post by swirlwood)
em..im sorry, but i really dont get that
You know the 5 SUVAT formulas? If not look in the back of your M1 book. Pick good values to allow you to solve a simultaneous equation. Hint: chose a equation that uses displacement time and acceleration.
7. ive tryd everythn i can fink of...i jus ned someone 2 possibly gt the ansa 4 me.?
8. (Original post by Mr M)
Right, protege know the answers and will look after you.

Do you know the equation ?
yepp i doo =D
9. (Original post by Pr0tégé)
You know the 5 SUVAT formulas? If not look in the back of your M1 book. Pick good values to allow you to solve a simultaneous equation. Hint: chose a equation that uses displacement time and acceleration.
Do that.
10. (Original post by Pr0tégé)
Do that.
..enm i did-unsuccessful
11. (Original post by swirlwood)
..enm i did-unsuccessful
You've probably forgotten how to do simultaneous equations. Make it so that when you minus the equations a value (u or a) cancels out.
12. s=ut+0.5at^2

Therefore, For AB:

80=4u+8a (1)

For AC:

140=6u+18a (2)

From equation (1),

40=2u+4a (multiply (1) by 0.5)
Therefore, 120=6u+12a (3)

(2)-(3) gives 20=6a
a=10/3 metres per second squared

Therefore substitute that into (1):

u= 40/3 metres per second

13. suvat!
14. (Original post by malolis)
s=ut+0.5at^2

Therefore, For AB:

80=4u+8a (1)

For AC:

140=6u+18a (2)

From equation (1),

40=2u+4a (multiply (1) by 0.5)
Therefore, 120=6u+12a (3)

(2)-(3) gives 20=6a
a=10/3 metres per second squared

Therefore substitute that into (1):

u= 40/3 metres per second

Thanks for telling him/her the answer.
15. ah thankyou, kind of got that now. thanks a lot...and as for language..sorry but really cant b asked when im stuck and stressed with maths...grr.
16. (Original post by malolis)
s=ut+0.5at^2

Therefore, For AB:

80=4u+8a (1)

For AC:

140=6u+18a (2)

From equation (1),

40=2u+4a (multiply (1) by 0.5)
Therefore, 120=6u+12a (3)

(2)-(3) gives 20=6a
a=10/3 metres per second squared

Therefore substitute that into (1):

u= 40/3 metres per second

sorry but im confusd with this part :40=2u+4a (multiply (1) by 0.5)
Therefore, 120=6u+12a (3)
why multiply by 0.5? if its for the sake of simplification...then why not divide by 4?
17. (Original post by swirlwood)
sorry but im confusd with this part :40=2u+4a (multiply (1) by 0.5)
Therefore, 120=6u+12a (3)
why multiply by 0.5? if its for the sake of simplification...then why not divide by 4?
How are you going to cancel a term if you divide by 4? Notice that there is a 6u in (2) and a 6u in (3).

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