A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car arrives at the position of the stop-light 7.5 s after the light had turned green.
What is the slowest constant speed which the green car can maintain and still catch up to the blue car?
Answer in units of m/s.
if a 3,000 kg car is moving at 10m/s and hits a 5,000 kg car at rest, what would be the resulting speed
To solve this problem, we need to determine the distance covered by the blue car in 7.5 seconds and find the slowest constant speed the green car can maintain in order to catch up.
First, let's calculate the distance covered by the blue car in 7.5 seconds using the equation:
distance = initial velocity * time + 0.5 * acceleration * time^2
The initial velocity of the blue car is 0 m/s (as it starts from rest), the time is 7.5 s, and the acceleration is 0.2 m/s^2. Plugging these values into the equation, we get:
distance = 0 * 7.5 + 0.5 * 0.2 * (7.5)^2
Simplifying the equation:
distance = 0.5 * 0.2 * 56.25
distance = 5.625 meters
So, the blue car covers a distance of 5.625 meters in 7.5 seconds.
To calculate the minimum speed the green car needs to maintain, it must cover this same distance of 5.625 meters in 7.5 seconds.
speed = distance / time
Plugging in the values:
speed = 5.625 / 7.5
speed = 0.75 m/s
Therefore, the slowest constant speed the green car can maintain and still catch up to the blue car is 0.75 m/s.
To find the slowest speed at which the green car can catch up to the blue car, we need to determine the position of the blue car when the green car arrives at the stop-light.
First, let's find the position of the blue car when the green car arrives at the stop-light. We know that the green car arrives 7.5 seconds after the light turned green. Therefore, the blue car has been accelerating for 7.5 seconds.
To find the position of the blue car, we will use the kinematic equation:
\(d = v_0 t + \frac{1}{2} a t^2\)
where
\(d\) is the position of the car,
\(v_0\) is the initial velocity of the car (in this case, 0 m/s because the car starts from rest),
\(a\) is the acceleration of the car (0.2 m/s\(^2\)),
and \(t\) is the time elapsed (7.5 s).
Plugging in the values, we have:
\(d = 0 \times 7.5 + \frac{1}{2} \times 0.2 \times (7.5)^2\)
\(d = 0 + 0.5 \times 0.2 \times 56.25\)
\(d = 0.5 \times 0.2 \times 56.25\)
\(d = 0.5 \times 11.25\)
\(d = 5.625\) meters
So, when the green car arrives at the stop-light, the blue car is 5.625 meters ahead.
Now let's find the slowest speed at which the green car can catch up to the blue car. By the time the green car reaches the stop-light, it needs to cover the same distance as the blue car from its starting position.
Let the slowest speed of the green car be \(v\) m/s.
Using the equation \(d = vt\), we can now find the time it takes for the green car to reach the stop-light:
\(5.625 = v \times 7.5\)
Rearranging the equation to solve for \(v\), we have:
\(v = \frac{5.625}{7.5}\)
\(v = 0.75\) m/s
Therefore, the slowest constant speed at which the green car can maintain and still catch up to the blue car is 0.75 m/s.