A car starts from rest and accelerates unifomly at 3m/s^2. A second car starts from rest 6 seconds later at the same point and accelerates uniformly at 5m/s^2. How long does it take the second car to over take the first car?
To find out how long it takes for the second car to overtake the first car, we can set up equations for their respective positions as functions of time.
Let's call the time when the second car overtakes the first car as "t".
For the first car:
The initial velocity (u1) is 0 m/s (starting from rest).
The acceleration (a1) is 3 m/s^2.
Using the equation of motion
s1 = u1*t + (1/2)*a1*t^2
For the second car:
The initial velocity (u2) is 0 m/s (starting from rest).
The acceleration (a2) is 5 m/s^2.
Using the equation of motion:
s2 = u2*(t-6) + (1/2)*a2*(t-6)^2
Since the distance traveled by both cars when the second car overtakes the first car is the same, we can equate these two equations:
u1*t + (1/2)*a1*t^2 = u2*(t-6) + (1/2)*a2*(t-6)^2
Substituting the given values:
0*t + (1/2)*3*t^2 = 0*(t-6) + (1/2)*5*(t-6)^2
Simplifying the equation:
(1/2)*3*t^2 = (1/2)*5*(t-6)^2
Now we can solve this quadratic equation for t.
To find out how long it takes for the second car to overtake the first car, we need to determine the time at which their displacements are equal.
Let's first calculate the displacement of the first car at any given time. Since it starts from rest and accelerates uniformly, we can use the formula:
s1 = (1/2) * a1 * t^2,
where s1 is the displacement of the first car, a1 is its acceleration (3 m/s^2), and t is the time.
For the second car, we need to consider the time difference of 6 seconds between their starts. Therefore, we can express the displacement of the second car as:
s2 = (1/2) * a2 * (t - 6)^2,
where s2 is the displacement of the second car, a2 is its acceleration (5 m/s^2), and (t - 6) is the time it takes for the second car to catch up.
Now, to find the time at which they meet, we set s1 = s2 and solve for t:
(1/2) * 3 * t^2 = (1/2) * 5 * (t - 6)^2.
Let's simplify this equation:
3t^2 = 5(t - 6)^2.
Expand the right side:
3t^2 = 5(t^2 - 12t + 36).
Now, distribute on the right side:
3t^2 = 5t^2 - 60t + 180.
Rearrange to bring everything to one side:
0 = 2t^2 - 60t + 180.
Divide both sides by 2 to simplify:
0 = t^2 - 30t + 90.
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Factoring in this case is not straightforward, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = 1, b = -30, and c = 90.
Plugging in these values:
t = (-(-30) ± √((-30)^2 - 4 * 1 * 90)) / (2 * 1).
Simplifying inside the square root:
t = (30 ± √(900 - 360)) / 2.
t = (30 ± √540) / 2.
Now, let's simplify the square root:
t = (30 ± √(36 * 15)) / 2.
t = (30 ± 6√15) / 2.
This can be further simplified as:
t = 15 ± 3√15.
So, the possible solutions for t are:
t1 ≈ 15 + 3√15,
t2 ≈ 15 - 3√15.
Both cars start at rest, so time cannot be negative. Therefore, the second car overtakes the first car approximately at t ≈ 15 + 3√15 seconds.
Solve this equation for t:
(3/2) t^2 = (5/2)(t-6)^2
You should get an equation of the form
at^2 + bt + c = 0.
Take the root that is t>6