Question

A particle of mass 2.0 kg is acted on by a single force →F1=18ˆiN. (a) What is the particle’s acceleration? (b) If the particle starts at rest, how far does it travel in the first 5.0 s?

a) (a) ( 9.0 , {m/s}^2 ), (b) 112.5 m
b) (a) ( 8.0 , {m/s}^2 ), (b) 100.0 m
c) (a) ( 10.0 , {m/s}^2 ), (b) 125.0 m
d) (a) ( 7.0 , {m/s}^2 ), (b) 87.5 m

Answer 1

The particle's acceleration is 9.0 ˆi m/s^2. The particle travels a distance of 112.5 m in the first 5.0 s.

Step-by-step explanation:

(a) To find the particle's acceleration, we need to use Newton's second law, which states that the force on an object is equal to its mass multiplied by its acceleration (F = ma). In this case, the force is given as 18ˆiN and the mass is 2.0 kg. So, we can calculate the acceleration using the formula: a = F/m = (18ˆiN)/(2.0 kg) = 9.0 ˆi m/s^2.

(b) If the particle starts at rest, we can use the kinematic equation s = ut + (1/2)at^2 to find the distance it travels in the first 5.0 s. Here, u represents the initial velocity, which is 0 m/s. Plugging in the values, we get: s = (1/2)(9.0 ˆi m/s^2)(5.0 s)^2 = 112.5 m. So, the correct answer is (a) (9.0, {m/s}^2), (b) 112.5 m.