A 3000kg car crashed to a 180kg motorcycle at rest, leading it to accelerate at 5m/s2. The car was pushed 15m back, how much farther is the motorcycle from the crash site? What is the final velocity of the car upon being pushed back that is caused by the crash?
To solve this problem, we can apply the principles of Newton's second law of motion and conservation of momentum.
Let's denote the initial velocity of the car as Vc (unknown) and the final velocity of the car as Vcf (also unknown). The initial velocity of the motorcycle is 0 m/s, and its final velocity is unknown. We also know the mass of the car (3000 kg) and the mass of the motorcycle (180 kg). The acceleration of the motorcycle, caused by the collision, is given as 5 m/s^2, and the displacement of the car is 15 m.
Let's start by calculating the velocity change of the car using the concept of conservation of momentum:
According to the principle of conservation of momentum, the sum of the initial momenta of the car and motorcycle should be equal to the sum of their final momenta. Mathematically, this can be written as:
(mass of car x initial velocity of car) + (mass of motorcycle x initial velocity of motorcycle) = (mass of car x final velocity of car) + (mass of motorcycle x final velocity of motorcycle)
(3000 kg x Vc) + (180 kg x 0 m/s) = (3000 kg x Vcf) + (180 kg x 5 m/s)
3000Vc = 3000Vcf + 900
Next, let's calculate the final velocity of the car (Vcf):
We know that the car moves in the opposite direction with respect to its initial velocity (Vc). Therefore, the final velocity of the car (Vcf) can be calculated using the equation of motion:
Vcf^2 = Vc^2 - 2ad
where:
Vcf = final velocity of car
Vc = initial velocity of car
a = acceleration
d = displacement
Substituting the given values, we get:
Vcf^2 = Vc^2 - 2(5 m/s^2)(15 m)
Now, we can calculate the final velocity of the car (Vcf):
Vcf^2 = Vc^2 - 150 m^2/s^2
Finally, to find the farther distance of the motorcycle from the crash site:
Since the car is pushed back by 15 m, the motorcycle will be at a distance equal to this displacement, plus any additional distance covered by the motorcycle due to its acceleration.
Thus, the motorcycle will be 15 meters plus the displacement covered in the extra time it takes to accelerate, given by the equation:
d = 15 m + (1/2)(5 m/s^2)(t^2)
where:
d = distance covered by the motorcycle
t = time taken for the acceleration
However, since we don't have the value of the time taken for acceleration (t) or the final velocity of the car (Vcf), we can't calculate the exact distance at this point.
To find the final velocity of the car and the farther distance of the motorcycle, we would need the missing information, such as the initial velocity of the car or the time taken for acceleration.