A particle of mass 5 kg is acted upon by a horizontal force of magnitude 16 N for 8 seconds, in which time it moves from rest until it is traveling with a speed v m/s. The particle continues to move with this speed for the next 15 seconds. it is brought to rest by the application of a constant resisting force of magnitude Y newtons. The total distance travelled is 1000 m.
2.Find v.
3. what is the total distance travelled while accelerating and traveling at a constant speed?
4. what is the rate at which the speed if the particle decreases before it comes to rest?
5.Find Y.
A "rocket car" is launched along a long straight track at = 0 . It moves with constant acceleration = 1.7 . At = 2.2 , a second car is launched with constant acceleration = 8.7 .
To find the answers to the questions, we need to use the equations of motion and apply Newton's second law.
1. To find the acceleration of the particle, we can use Newton's second law: F = ma, where F is the force applied to the particle, m is the mass, and a is the acceleration.
Given: F = 16 N, m = 5 kg.
Using F = ma, we can solve for a:
16 N = 5 kg * a
a = 16 N / 5 kg
a = 3.2 m/s^2
2. To find the final velocity (v), we need to use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since it starts from rest), a is the acceleration, and t is the time elapsed.
Given: u = 0 m/s, a = 3.2 m/s^2, t = 8 s.
Using v = u + at, we can solve for v:
v = 0 m/s + 3.2 m/s^2 * 8 s
v = 25.6 m/s
Therefore, the value of v is 25.6 m/s.
3. To find the total distance traveled during acceleration and at a constant speed, we need to calculate the distance traveled during acceleration and the distance traveled at a constant speed, and then add them together.
Distance traveled during acceleration can be calculated using the equation: s = ut + (1/2) a t^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time elapsed.
Given: u = 0 m/s, a = 3.2 m/s^2, t = 8 s.
Using s = ut + (1/2) a t^2, we can solve for the distance traveled during acceleration:
s1 = 0 m/s * 8 s + (1/2) * 3.2 m/s^2 * (8 s)^2
s1 = 0 m + 0.5 * 3.2 m/s^2 * 64 s^2
s1 = 102.4 m
Distance traveled at a constant speed can be calculated using the equation: s = vt, where s is the distance traveled, v is the constant velocity, and t is the time elapsed.
Given: v = 25.6 m/s, t = 15 s.
Using s = vt, we can solve for the distance traveled at a constant speed:
s2 = 25.6 m/s * 15 s
s2 = 384 m
Total distance traveled = s1 + s2 = 102.4 m + 384 m = 486.4 m
Therefore, the total distance traveled while accelerating and at a constant speed is 486.4 m.
4. The rate at which the speed of the particle decreases before it comes to rest is equal to the magnitude of the resisting force. Since the particle is brought to rest by a constant resisting force of magnitude Y newtons, the rate of decrease in speed is Y newtons.
5. To find the value of Y, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity (which is 0 since the particle comes to rest), u is the initial velocity, a is the acceleration, and s is the distance traveled.
Given: v = 0 m/s, u = 25.6 m/s, s = 1000 m.
Using v^2 = u^2 + 2as, we can solve for Y:
(0 m/s)^2 = (25.6 m/s)^2 + 2 * a * 1000 m
0 = (25.6 m/s)^2 + 2 * 3.2 m/s^2 * s
0 = 655.36 m^2/s^2 + 6.4 m/s^2 * 1000 m
0 = 655.36 m^2/s^2 + 6400 m^2/s^2
Y = least magnitude such that the equation is satisfied
Therefore, to find the value of Y, we need to solve the equation and find the least magnitude so that the equation is satisfied.