Annoying physics problem?

Based on the given condition of the problem, it will be ASSUMED that after reaching max velocity, each person will then be running the rest of the distance without any acceleration. In other words, the runners will only be accelerating until they reach their maximum speeds.

Working formula is

S = (1/2)AT^2 + V(10.7 - T)

where

S = 100 meters

A = acceleration of the runner

T = time when runner reaches maximum speed

V = maximum speed

For Person A,

100 = (1/2)Aa(4) + Va(10.7 - 2)

100 = 2Aa + 8.7Va --- call this Equation A

Also,

Aa = (Va - 0)/2 = Va/2 and substituting this into Equation A,

100 = Va + 8.7Va

9.7Va = 100

Va = 10.31 m/sec.

and Person A's acceleration is

Aa = 10.31/2 = 5.16 m/sec^2

For Person B, the calculation procedure is basically the same as the above procedure for Person A,

100 = (1/2)(Ab)(3)^2 + Vb(10.7 - 3)

100 = 4.5Ab + 7.7Vb --- call this Equation B

also,

Ab = Vb/3

and substituting this into Equation B,

100 = 1.5Vb + 7.7Vb

and solving for Vb,

Vb = 100/9.2

Vb = 10.87 m/sec.

and solving for Ab,

Ab = 10.87/3 = 3.62 m/sec^2

SUMMARY --

For Person A : Va = 10.31 m/sec and Aa = 5.16 m/sec^2

For Person B : Vb = 10.87 m/sec and Ab = 3.62 m/sec^2

<< At 6 seconds what are the positions of both? >>

For Person A,

Da = (1/2)(5.16)(2)^2 + (10.31)(6 - 2)

Da = 51.56 m

For Person B,

Db = (1/2)(3.62)(3)^2 + 10.87(6 - 3)

Db = 48.90 m

Hope this helps.

Assuming constant acceleration, you have that the average velocity during a period of time is (v0 + v1)/2 -- just the average of the endpoints. If v0 is 0, and v1 = v, then this is just v/2.

The total time of the dash is T (10.7 s). The time of acceleration is t. Then the distance travelled is:

[ t * (v/2) ] + [ (T - t) * v ] = 100 m.

This actually lets you solve directly for v, since:

v [ t/2 + T - t ] = 100 m

v [ T + t/2 ] = 100 m

v = 100 m / [ T + t/2 ]

The rate of acceleration is gotten by realizing that you go from 0 to v in t seconds: therefore, it is (v - 0)/t = v/t.

The position after 6s is gotten similarly to how we just got the position before, except with 6s instead of T:

x = [ t * (v/2) ] + [ (6s - t) * v ]

laws used:

s=ut+0.5at^2

v=u+at

equations:

1. 2a1+x=100......x is the distance covered at max velocity, 2a1 is distance covered during accelaration for person A using law 1

2. 9/2a2 + y=100 .....same for person B

3. x=v1(max) * 8.7

4. y=v2(max) * 7.7

5. v1(max) = 2a1

6. v2(max) = 3a2

put 3, 4 ,5, 6, in 1 and 2 to get v1max and v2max, solve the rest of the problem using these equations

Let u, v, w be the speeds at time zero, a million.five and a couple of.6 respectively. Distance traveled = usual speed x time eight.6 = [v +u] t1 / two [v +u] = two x eight.6 / a million.five= eleven.forty seven --------A [w +v] = two x eight.6 /a million.a million=15.sixty three--------------B Subtracting A from B [w - u]= four.sixteen Acceleration a = [w - u] / [time to difference from u to w] a = four.sixteen / two.6 = a million.6 m/s^two ===================================== Or The usual pace for the primary automobile U = eight.6 / a million.five m/s For the moment one V = eight.6 / a million.a million m/s. Using V^two - U^two = 2as a = [V^two - U^two] / two * eight.6 = a million.sixty four m/s^two