Solving for velocity,etc..

Question

an automobile moving at a constant velocity of 18m/s passes a gasoline station. 3 sec later, another automobile leaves the same gasoline station and accelerates from rest at a rate of 1.9m/s^2.

A. when and where will the two vehicles meet?

B. what are the speed when they are side by side?

Answer 1

Xa=VaXt1,Xb=1/2(1.9){(t1+3)^}2,equqting both equal;18t1=0.95 X{t1^2+6t1}=0.95t1^2+8.55+5.7t1,Solving for t1 you get t1=12.21 & 1.4 seconds.Now we must put these two quantities in our original equations;18X12.21=219.78 and Xb=219.78 o this answer is O.K.

for 1.4 sec,Xa=18X1.4=25.2 and Is NOT accepted because it is less than 3.So in a distance of 219.78 meter from the station they meet,15.21 seconds after the the second car leaves the station.

Va is constant so it is 18m/sec,Vb=0.95X15.21=14.4495 m/sec.

Answer 2

Well the first car's position will be described by the equation:

y_1 = 18x + 54   x is in seconds

And the second car's position will be:

y_2 = 1.9(x^2) + 0

so just set the two equal

18x + 54 = 1.9x^2  or   1.9x^2 - 18x - 54 = 0

(you need a calculator to find the zeros)

That gives you time for A; plug in the time to one of the equations to get the distance

For B, the first car is still going 18 m/s, so just take the time (T) and multiply it by the acceleration of car 2 and that will give you its speed

1.9 x T

Hope this helps

Answer 3

If d the distance when they meet.

1st Automobile:

d = vt = 18t

2nd Automobile:

d = (1/2)a(t - 3)² = 0.5*1.9*(t -3)²

Equaling the equations:

18t = 0.5*1.9*(t -3)²

18.947t = t² - 6t + 9

t² - 24.947t + 9 = 0

Answer A:

t = 24.58 sec

d = 24.58*18 = 442.44 m

Answer B:

Speed of the 2nd automobile:

v = at

v = 1.9*24.58 = 46.7 m/s

Good luck.

Nozar nazari: Xb=1/2(1.9){(t1+3)^}2 is not correct

2nd Automobile to uses (t-3) sec.