An automobile moving at a constant velocity of 15m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 2m/s^2 in the same direction. How soon does the second autombile overtake the first?
(15*2)+(15t)=(1\2)(2)t^2
By solving this quadratic equation
t=16.7sec
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To solve this problem, we need to determine the time it takes for the second automobile to catch up with or overtake the first automobile.
Let's break down the information we have:
1. Initial velocity of the first automobile (v1): 15 m/s
2. Acceleration of the second automobile (a2): 2 m/s^2
3. Time delay before the second automobile starts (t): 2 seconds
First, we need to find out the position of the first automobile at the time the second automobile starts moving. We can calculate it using the equation:
s1 = v1 * t
where s1 is the position of the first automobile.
s1 = 15 m/s * 2 s
s1 = 30 m
So, the first automobile has traveled 30 meters by the time the second automobile starts.
Next, we can calculate how long it takes for the second automobile to catch up with the first. We can use the following equation, known as the motion equation:
s2 = s1 + v2 * t + (1/2) * a2 * t^2
where s2 is the position of the second automobile, v2 is its velocity at time t, and a2 is its acceleration.
Since the second automobile starts from rest (0 m/s), we can rewrite the equation as:
s2 = s1 + (1/2) * a2 * t^2
Substituting the values we have:
s2 = 30 m + (1/2) * 2 m/s^2 * t^2
Now, we want to find the time (t) when the second automobile catches up with the first. Rearranging the equation gives us:
t^2 = (2 * (s2 - s1)) / a2
t^2 = (2 * (s2 - 30 m)) / 2 m/s^2
t^2 = s2 - 30 m
Now we can solve for t by substituting the relationship between the position of the second automobile and its velocity:
s2 = v2 * t
s2 = 0 m/s * t (as v2 = 0 initially since the second car starts from rest)
s2 = 0 m
Substituting s2 = 0 m into the equation gives:
t^2 = 0 m - 30 m
t^2 = -30 m
We can see that the equation has a negative value, which does not make sense in this context. Therefore, we can conclude that the second automobile does not overtake the first automobile.
Hence, the second automobile does not overtake the first automobile in this scenario.
Let t =0 be the time after the first can passes by.
Solve for t when
15 t = (1/2)*(2.0)*(t-2)^2
The term on the right is the distance covered by the accelerating car that starts 2 seconds late.
15t = (t-2)^2 = t^2 -4t + 4
t^2 -19t +4 = 0
t = [19 +sqrt(345)]/2 = 18.8 seconds
They both will be 282 m from the gas station at that time.