Two cars pass each other traveling at the same speed. One car has a constant velocity of 18.0 m/s east. The other car has a constant acceleration of 3.00 m/s^2, west. How much time will have elapsed until the cars are 128 m apart?
To find the time it takes for the cars to be 128 m apart, we can use the equation of motion for the car with constant acceleration.
The equation is given as: s = ut + (1/2)at^2
Where:
- s is the distance traveled
- u is the initial velocity
- t is the time elapsed
- a is the acceleration
Let's find the time it takes for the car with constant velocity to travel 128 m:
s = ut
128 m = 18.0 m/s * t
Now let's find the time it takes for the car with constant acceleration to travel 128 m:
s = ut + (1/2)at^2
128 m = 0 * t + (1/2) * 3.00 m/s^2 * t^2
Since the two cars are traveling in opposite directions, the total distance traveled by both cars should be equal to 128 m. Therefore, the sum of the distances traveled by each car should be 128 m.
Let's set up the equation:
18.0 m/s * t + (1/2) * 3.00 m/s^2 * t^2 = 128 m
Simplifying the equation:
9.0 m/s * t + 1.5 m/s^2 * t^2 = 128 m
1.5 t^2 + 9.0 t - 128 = 0
To solve this quadratic equation for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = 1.5, b = 9.0, and c = -128.
Calculating the values:
t = (-9.0 ± √(9.0^2 - 4 * 1.5 * -128)) / (2 * 1.5)
Simplifying further:
t = (-9.0 ± √(81 + 768)) / 3
t = (-9.0 ± √849) / 3
Using the positive value:
t = (-9.0 + √849) / 3
Now we can calculate t using a calculator:
t ≈ 4.60 seconds
Therefore, it would take approximately 4.60 seconds for the cars to be 128 meters apart.
To find the time it will take for the cars to be 128 m apart, we can use the equation of motion:
d = v0t + (1/2)at^2
In this equation, d is the distance traveled, v0 is the initial velocity, t is the time, and a is the acceleration.
For the first car with constant velocity, we have:
d1 = v0t1
Since the velocity is constant and equal to 18.0 m/s, we have:
d1 = 18.0 t1
For the second car with constant acceleration, we have:
d2 = v0t2 + (1/2)at2^2
Since the initial velocity is 0 m/s (as it starts from rest) and the acceleration is -3.00 m/s^2 (opposite direction), we have:
d2 = (1/2)(-3.00)(t2^2) = -1.50 t2^2
The total distance between the cars is the sum of these distances:
d1 + d2 = 18.0 t1 - 1.50 t2^2
Since the cars are traveling in opposite directions, the total distance is equal to 128 m:
18.0 t1 - 1.50 t2^2 = 128
Now, we need to find the relationship between t1 and t2. Since both cars are traveling at the same speed, their distances traveled will be equal when they meet. Therefore, we can say:
d1 = d2
18.0 t1 = -1.50 t2^2
Simplifying this equation, we have:
t1 = -1.50 t2^2 / 18.0
Substituting this value of t1 in the equation 18.0 t1 - 1.50 t2^2 = 128, we can solve for t2:
18.0 (-1.50 t2^2 / 18.0) - 1.50 t2^2 = 128
Simplifying further:
-1.50 t2^2 - 1.50 t2^2 = 128
-3.00 t2^2 = 128
Dividing both sides by -3.00:
t2^2 = -128 / -3.00
t2^2 = 42.6667
Taking the square root of both sides to solve for t2:
t2 = √42.6667
t2 = 6.528 m/s
Therefore, it will take approximately 6.528 seconds for the cars to be 128 m apart.
s1=v•t
s2= v•t+a•t²/2
180= s1+s2 = v•t +v•t+a•t²/2=2 v•t+a•t²/2
a•t² +4 v•t -360 = 0
3 t²+72t-360 =0
t²+24t-120 =0
1= -12±sqrt(144+120)
t=4.25 s.