A truck is traveling at 16 m/s to the north. The driver of a car, 500 m to the north and traveling south at 25 m/s, puts on the brakes and slows at 3.5 m/s2. Where do they meet?
please help!!
Let the truck position be X1 and the car position X2, measured along the same north-south line. Let x = 0 be the point where the truck is when the car starts braking.
X1 = 16 t
X2 = 500 - 25 t - 1.75 t^2
They meet when X1 = X2. Solve for t at that time.
To find the time when they meet, we need to set X1 equal to X2 and solve for t.
16t = 500 - 25t - 1.75t^2
First, rearrange the equation to get a quadratic equation in standard form:
1.75t^2 + 41t - 500 = 0
To solve this equation, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1.75, b = 41, and c = -500. Plugging in these values, we get:
t = (-41 ± √(41^2 - 4 * 1.75 * -500)) / (2 * 1.75)
Now, calculate the value inside the square root:
√(41^2 - 4 * 1.75 * -500) = √(1681 + 3500) = √5181 = 71.96
Plug this back into the equation:
t = (-41 ± 71.96) / 3.5
Now we have two possible values for t:
t1 = (-41 + 71.96) / 3.5 ≈ 8.27
t2 = (-41 - 71.96) / 3.5 ≈ -34.27
Since time (t) cannot be negative in this context, we discard t2.
Therefore, the two vehicles meet approximately at t1 ≈ 8.27 seconds.
To find the position where they meet, plug this value of t back into either X1 or X2 equation:
X = 16 * t = 16 * 8.27 ≈ 132.32 meters
Therefore, the truck and the car will meet approximately 132.32 meters to the north of the starting point of the truck.
To find where they meet, set X1 equal to X2:
16t = 500 - 25t - 1.75t^2
Rearrange the equation to form a quadratic equation:
1.75t^2 + 41t - 500 = 0
Now we can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1.75, b = 41, and c = -500. Substituting these values into the quadratic formula:
t = (-41 ± √(41^2 - 4 * 1.75 * -500)) / (2 * 1.75)
Simplifying further:
t = (-41 ± √(1681 + 3500)) / 3.5
t = (-41 ± √5181) / 3.5
Taking the positive root:
t = (-41 + √5181) / 3.5 ≈ 6.73 seconds
Now, substitute the value of t back into one of the original equations to find the position where they meet. Using X2:
X2 = 500 - 25 * 6.73 - 1.75 * 6.73^2 ≈ 268.79 meters
Therefore, they meet approximately 268.79 meters to the north of the starting point of the car (or 500 - 268.79 ≈ 231.21 meters to the south of the starting point of the truck).