A race car is moving at a constant speed of 35 meter per second. as the race car passes by it and was accelerating at a constant rate of 5meter per second square .what was the speed of the security car when it took over the race car?
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To find the speed of the security car when it took over the race car, we need to determine the time it takes for the security car to catch up with the race car.
First, let's find the time it takes for the security car to catch up with the race car. We can use the equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
Since the race car is moving at a constant speed, its initial velocity is 35 m/s and its acceleration is 0. The distance it travels is the same as the distance traveled by the security car, so we can rewrite the equation as:
distance = security car's initial velocity × time + 0.5 × acceleration × time^2
We want to find the time it takes for the security car to catch up with the race car, so we need to rearrange the equation to solve for time:
0.5 × acceleration × time^2 + security car's initial velocity × time - distance = 0
Now, we can substitute the given values into the equation. Since the race car and security car start at the same initial position, the distance they both travel is 0.
0.5 × 5 × time^2 + security car's initial velocity × time - 0 = 0
Simplifying the equation:
2.5 × time^2 + security car's initial velocity × time = 0
Since we know the acceleration and distance are both positive, we can disregard the negative solution.
Now, we need to find the initial velocity of the security car when it starts catching up to the race car. The race car is moving at a constant speed of 35 m/s.
To find the initial velocity of the security car, we need to find the time it takes for the security car to catch up with the race car. Plugging in the given values, the equation becomes:
2.5 × time^2 + security car's initial velocity × time = 0
Now we have a quadratic equation. To solve for time, we can use the quadratic formula:
time = [-b ± sqrt(b^2 - 4ac)] / 2a
In this case, a = 2.5, b = security car's initial velocity, and c = 0. Plugging these values into the quadratic formula, the equation becomes:
time = [-security car's initial velocity ± sqrt((security car's initial velocity)^2 - 0)] / 2(2.5)
Since time cannot be negative, we can disregard the negative solution.
Now, we know the time it takes for the security car to catch up with the race car. We can substitute this value back into the equation we derived earlier:
distance = security car's initial velocity × time + 0.5 × acceleration × time^2
Since the distance is 0, the equation becomes:
0 = security car's initial velocity × time + 0.5 × acceleration × time^2
Plugging in the values of the security car's acceleration and the time it takes, the equation becomes:
0 = security car's initial velocity × time + 0.5 × 5 × time^2
Now, we can solve for the initial velocity of the security car:
35 = security car's initial velocity × time + 0.5 × 5 × time^2
35 = security car's initial velocity × time + 2.5 × time^2
Since we know the time it takes, we can substitute the value back into the equation:
35 = security car's initial velocity × time + 2.5 × (time)^2
Substituting the value of time, we get:
35 = security car's initial velocity × time + 2.5 × (time)^2
Now, we can solve for the initial velocity of the security car by substituting the value of time:
35 = security car's initial velocity × (time) + 2.5 × (time)^2
Now, we can simplify the equation and solve for the initial velocity of the security car.
It will take t seconds, which is when
35t = 5/2 t^2
The speed is then 5t