An automobile moving at a constant velocity of 14 m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 1.8 m/s². How soon will the second automobile overtake the first?
14 * t = (1.8 / 2) t^2
d1 = r*T = 14*2 = 28 m.
d2 = 0.5*a*T^2.
0.5*a*T^2 = 28 +14T
0.5*1.8T^2 = 28+14T.
0.9T^2-14T-28 = 0
T = ( -B+-sqrt(B^2-4AC))2A
T = (14+-sqrt(196+100.8))/1.8 = (14+-17.2)/1.8 = 17.3 s, and -1.78 s.
T = 17.3 s.
To solve this problem, we will use the equation of motion for the second automobile:
\[ s = ut + \frac{1}{2}at^2 \]
Where:
- \( s \) is the distance covered by the second automobile
- \( u \) is the initial velocity of the second automobile
- \( a \) is the acceleration of the second automobile
- \( t \) is the time taken by the second automobile
For the first automobile, we know that it is moving at a constant velocity of 14 m/s. The distance it covers after 2 seconds can be calculated as:
\[ s = ut \]
\[ s = 14 \times 2 \]
\[ s = 28 \, \text{m} \]
Now, let's determine when the second automobile overtakes the first. We can set the equations for the distance covered by the first and second automobiles equal to each other:
\[ s_1 = s_2 \]
\[ 28 + 14t = 1.8t^2 \]
Rearranging the equation, we get:
\[ 1.8t^2 - 14t - 28 = 0 \]
To solve this quadratic equation, we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
By substituting the values, we can find the value of \( t \).