A particle of mass 1kg moving with initial velocity (i+2k)m/s is acted upon by a constant force(i+2j-2k)N,calculate the velocity and distance after 5s.
a = F/1 = i + 2 j -2 k
v = Vi + a t
x = Xi + Vi t + (1/2)a t^2
v at 5 = i+0j+2k + 5i +10j -10k
= 6i + 10j - 8k
x at 5, well assume start at (0,0,0)
x = 0 +(i+0j+2k)5 +(1/2)(i + 2 j -2 k)25
= 5i+0j+10k+(1/2)(25i+50j-50k)
=17.5 i +25 j -15 k
check my arithmetic !
To determine the velocity and distance after 5 seconds, we can use the equations of motion from Newton's second law of motion.
First, let's calculate the acceleration (a) using the formula:
F = ma
Given that the force (F) acting on the particle is (i + 2j - 2k) N and the mass (m) is 1 kg, we can substitute these values into the equation to find the acceleration:
(i + 2j - 2k) N = 1 kg * a
Therefore, a = (i + 2j - 2k) N.
Next, we can calculate the change in velocity (Δv) using the formula:
Δv = a * t
Given that the acceleration (a) is (i + 2j - 2k) N and the time (t) is 5 s, we can substitute these values into the equation to find the change in velocity:
Δv = (i + 2j - 2k) N * 5 s
Now, let's calculate the final velocity (v) by adding the change in velocity to the initial velocity:
v = (i + 2j - 2k) m/s + (Δv)
Given that the initial velocity is (i + 2k) m/s, we can substitute this value into the equation:
v = (i + 2k) m/s + (Δv)
Finally, to calculate the distance traveled (s), we can use the formula:
s = (initial velocity * time) + (0.5 * acceleration * time^2)
Given that the initial velocity is (i+2k) m/s, the time is 5 seconds, and the acceleration is (i + 2j - 2k) N, we can substitute these values into the equation to find the distance traveled:
s = (i+2k) m/s * 5 s + 0.5 * (i + 2j - 2k) N * (5 s)^2
By computing the above equations, you will obtain the values for the final velocity and distance after 5 seconds.