If two cars are traveling along a straight road, one behind the other. the first is traveling at a constant velocity of 10 m/s. the second, approaching from the rear, is traveling at 22 m/s. when the second car is 300m behind the first, the driver applies the brakes, producing a constant acceleration of -0.40 m/s^2. will the cars crash and if so, where and when?
I find this really hard to understand
will it hit?
use final velocity of 0 presuming that it came to a halt
find the time elasped for it to stop using acceleration = change in v over change in t
use the time to find how far both have traveled...
yes it will hit
it;s an equation of what time used for second car to travel 300+11t
since we don't know the final velocity cant use equations of it
so use change in x = intial v (t) + 1/2 a t^2
so 300+11t=22t+(-2)t^2
2t^2-11t+300=0
using solve t using quadratic formula to see what time will it hit
To determine if the cars will crash and if so, where and when, we need to calculate the time it takes for the second car to catch up to the first car.
Step 1: Find the initial separation between the cars.
The initial separation is given as 300 meters.
Step 2: Find the relative velocity between the two cars.
The relative velocity is the difference between the velocities of the two cars.
Relative velocity = Speed of the second car - Speed of the first car
Relative velocity = 22 m/s - 10 m/s
Relative velocity = 12 m/s (since the second car is approaching from the rear, the relative velocity is positive)
Step 3: Find the time it takes for the second car to catch up to the first car.
We can use the equation of motion to find the time:
Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2
Since the acceleration is constant, the equation simplifies to:
Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2
Substituting the known values:
300m = 12 m/s * Time + (1/2) * (-0.40 m/s^2) * Time^2
Simplifying the equation, we have a quadratic equation:
0.2 Time^2 - 12 Time + 300 = 0
To solve this equation, we can use the quadratic formula:
Time = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values:
Time = (-(-12) ± sqrt((-12)^2 - 4 * 0.2 * 300)) / (2 * 0.2)
Solving the equation gives us two possible times:
Time ≈ 47.32 seconds (ignoring the negative value)
So, it will take approximately 47.32 seconds for the second car to catch up to the first car.
Step 4: Determine if the cars will crash and where.
To determine if the cars will crash, we need to calculate the distance traveled by the first car during the calculated time.
Distance traveled by the first car = Velocity of the first car * Time
Distance traveled by the first car = 10 m/s * 47.32 s
Distance traveled by the first car ≈ 473.2 meters
Since the distance traveled by the first car is greater than the initial separation between the cars (300 meters), the cars will indeed crash.
The crash will occur at a distance of approximately 300 meters from the starting point of the first car.