Just A Quick Question To Solve

solve

initial Velocity = 50 Kilometres Per Hour final Velocity = 30 Kilometers Per Hour Distance covered = 200 Meters

Time = ? Deceleration (acceleration) = ?

Muhammad Umar Farooq

You can solve this only if you assume something else about the acceleration, for example that it is constant. Then v1-v0 = A t, s = v0 t + A/2 t^2

Two equations, two unknowns... Ciao -erk-

With constant acceleration, or not? (looking at the groups posted to, probably so)

s = 1/2 (u + v) t

v^2 = u^2 + 2 a s

Of course, if acceleration is not contstant, then you'll need some slighter higher order equations using more advanced calculus or similar

And don't forget to check your units...

OK, constant decelration 200 m = .2 Km your first equation goes like 0.2 = 1/2 (80) t t = 1/200 Hr = 18 sec your second equation goes like 900 = 2500 + 2 a 0.2 a = -4000 Km/Hr/Hr = -1111.11 m/s/s

I think so, Am I right ?

eliminate A from above two 50-30 = A t , = 20 0.2 = 50 t + (1/2) (A t) (t) 0.2 = 50 t + 20/2 t 0.2 = 60 t t = 1/300 Hr = 12 sec A t = 20 A = 20/(1/300) A = 6000 Km/Hr/Hr = 1666.67 m/s/s

compare my reply of Stephen's email t = 18 sec A = 1111.11 m/s/s where is the flaw ?

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How did you get the time in v=u+at ?? (when both a and t are unknown ). Only the second law of motion , ie., v^2 - u^2 = 2 * a * s is applicable. ( here v , u and s are known ).

Assuming constant decelaration.... Acceleration = -4000Km/hour ( -ve and hence decelaration ). Now time can be found using the first formula .

Therefore, time = 18 seconds..

Geee....Your answers are right though !!!!

Is it an African or an Asian swallow, and how many coconuts are they carrying? Is there a smell of elderberry wine or hamsters? Are you being oppressed?

No, Umar is right. In uniformly accelerated motion, you can actually use the average velocity, 1/2(u+v), to relate displacement and time:

s = 1/2(u+v)*t

So that lets you solve for time without needed a quadratic equation.

Any equations derived from

v = u + at s = u*t + 1/2*a*t^2

are correct.

You can prove the average velocity formula geometrically (the displacement is the area under the velocity 'curve', which is a trapezoid with sides u and v).

You could also do it algebraically by substituting a = (v-u)/t in the equation for s:

s = u*t + 1/2*t^2*(v-u)/t = u*t + 1/2*v*t - 1/2*u*t = 1/2(u+v)*t

- Randy

Well, if you think so, then you may well be right, after all, I'm not a teacher, nor a physics or maths graduate ...

I think the point Al was trying to make (unless I've misintrepeted his reply) was that it's generally 'bad form' to have to rely on calculated answers to plug into a later expression, if you can avoid it (at least in my book). Either that, or it maybe that we've misunderstood the

line as referring to v=u+at (when it refers to v^2=u^2+2as) ?

referring to v^2=u^2+2 a s (the second equation in your first email)