Ais running at 3.8m/s,75m behind B who is running at 4.2m/s.if A accelerate at 0.15m/s-2,how long Acancatch b?

Question

how long does it take for A to catch up B? and also wts the distance when A can catch up B?

Answer 1

We have to set up an equation. Since A is accelerating at 0.15 m/sec/sec, the time will be when A travels the 75 m extra.

3.8t + 1/2 0.15 t^2 = 75 + 4.2t

0.075t^2 - 0.4t - 75 = 0

Dividing by 0.075 we can rewrite the equation as

t^2 - 5.33t - 1000 = 0

This is a quadratic equation in t. Solving for t, we get

t1 = [- (-) 5.33 + sqrt.(5.33^2 + 4000)] / 2

and t2 = [ 5.33 - sqrt(5.33^2 + 4000)] / 2


t1 = [5.33 + 63.47] / 2 = 68.8/2 = 34.4 s

We can ignore t2 since it will be -ve and has no real significance.

Distance where they meet will be

75 + 4.2t = 75 + 4.2 x 34.4 = 219.48 m = 220 m approx.

Answer 2

dertermine the motion of both A and B

motion of A is an acceleration
motion of B is constant

Xf (A) = .5at^2 + V*t + Xi
Xf (B) = vt

Xf = final position
a = acceleration
t = time
V = intial velocity
Xi = initial position

A is behind 75m behind B, because A is accleration, at a certain distane and a certain time, A will catch up with B and their distances will be equal

Xf (A) = Xf (B)
.5at^2 + Vt + Xi = Vt

let Xi of B is 0m, then Xi of A is -75m
.5(.15)t^2 + 3.8t - 75 = 4.2t

0.075t^2 - 0.4t - 75 = 0

use quadratic fomula and you'll get t = 34.401681s

Xf = .5at^2 + Vt + Xi
Xf = .5(.15)(34.401681)^2 + (3.8)(34.401681) - 75
Xf = 144.487062m

144.48 is the position A ralative to the starting point of B. If you meant to find the total distance of A, then 144.48 + 75 = 219.48m is the answer

Answer 3

let they meet .
75+x =3.8t+.075t^2 [motion of A ]
x = 4.2t
solving we get
x = 144.48m
t = 38.40 time taken for catch up
hence A will meet after travelling 219.48 m from initial position of A