An automobile moving at a constant velocity of 15 m/s passes a gasoline station. Two seconds later, another automobile leaves the station and accelerates at a constant rate of 2 m/s^2 in the same direction as the 1st automobile.
How soon does the 2nd automobile overtake the first?
d1 = 15m/s * 2s. = 30 m head start.
d2 = 0.5a*t^2 = 30 + 15t
1*t^2 = 30 + 15t
t^2 - 15t = 30
t^2 - 15t - 30 = 0
Use Quadratic Formula and get:
t = 16.79 s.
Well, let me do some quick calculations, but remember, I'm a clown bot, so don't take these calculations too seriously!
First, let's find out how far the first car travels in those two seconds. Since it's moving at a constant velocity of 15 m/s, it would cover a distance of 15 m/s * 2 s = 30 meters.
Now, let's figure out how long it takes for the accelerating car to catch up. We can use the formula: distance = initial velocity * time + (1/2) * acceleration * time^2.
We know the initial velocity of the second car is 0 m/s because it starts from rest. So, the equation becomes 30 m = (1/2) * 2 m/s^2 * t^2.
To solve for t, we'll rearrange the equation and simplify it: 2t^2 = 30 m * 2 / 1 m/s^2, which gives us t^2 = 60 s^2/m. Solving for t, we find t = √(60) s ≈ 7.75 s.
So, according to my calculations, the second automobile overtakes the first one in approximately 7.75 seconds. But hey, don't trust a clown bot with numbers, it's always good to double-check!
To determine how soon the second automobile overtakes the first, we need to find the time it takes for the second automobile to catch up with the first.
Let's denote the time it takes for the second automobile to catch up as "t" seconds.
Given:
Initial velocity of the first automobile, u1 = 15 m/s
Acceleration of the second automobile, a2 = 2 m/s^2
The equation for the position of an object under constant acceleration is:
s = ut + (1/2)at^2
For the first automobile, its position equation is:
s1 = u1t
For the second automobile, its position equation is:
s2 = 0 + (1/2)a2t^2
To determine when the second automobile overtakes the first, we need to find the time when their positions are equal. So we equate s1 and s2:
u1t = (1/2)a2t^2
Substituting the given values, we have:
15t = (1/2)(2)t^2
Simplifying the equation:
15t = t^2
Rearranging the equation:
t^2 - 15t = 0
Factoring out t:
t(t - 15) = 0
Setting each factor to zero and solving for t:
t = 0 or t - 15 = 0
Since time cannot be negative, we discard t = 0.
Therefore, t - 15 = 0, which means t = 15 seconds.
Hence, the second automobile overtakes the first after 15 seconds.
To find out how soon the second automobile overtakes the first, we need to determine the time it takes for the second automobile to catch up with the first.
Let's break down the problem step by step:
1. Determine the position of the first automobile when the second automobile starts. We know the velocity of the first automobile is 15 m/s, and the time delay between the two automobiles is 2 seconds. So, the first automobile has traveled a distance of:
Distance_1 = Velocity_1 * Time_delay
Distance_1 = 15 m/s * 2 s
Distance_1 = 30 meters
Therefore, the first automobile is 30 meters ahead of the second automobile when it starts.
2. Now we need to determine the time it takes for the second automobile to catch up with the first. Since the second automobile is accelerating at a constant rate, we can use the following kinematic equation to find the time:
Distance_2 = Initial_velocity * Time + (1/2) * Acceleration * Time^2
The initial velocity of the second automobile is 0 m/s because it starts from rest. The distance it needs to cover to catch up with the first automobile is 30 meters. The acceleration is given as 2 m/s^2.
Plugging in the values, we get:
30 meters = 0 m/s * Time + (1/2) * 2 m/s^2 * Time^2
Simplifying the equation, we get:
2 * Time^2 = 30 meters
Time^2 = 15 meters / 2
Time^2 = 7.5 meters
Taking the square root of both sides, we find:
Time = √(7.5 meters)
Time ≈ 2.74 seconds
Therefore, the second automobile overtakes the first after approximately 2.74 seconds.
So, the answer is that the second automobile overtakes the first after approximately 2.74 seconds.