A particle of mass 5 kg is undergoing one-dimensional motion. It is subject to a constant force of 25 N, and has an initial speed of 3 m/s. The force is parallel to the initial velocity, both of which are directed in the positive direction along the axis of motion
(a) what is the change in velocity △v, in meters per second, of the particle between t = 0 and t = 2 s?
(b) What is the displacement △r, in meters, o ,i m, of the particle betwemt-0 and-2 t = 2 s?

Question

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lublana lublana · Answer 1 · 2020-02-13T10:42:03+01:00

Answer:

(a)10m/s

(b)16 m

Explanation:

We are given that

Mass of particle=m=5 kg

Force=F=25 N

Initial speed=u=3m m/s

a.t=0 and t=2 s

We know that

Force=[tex]F=ma[/tex]

Using the formula

[tex]25=5a[/tex]

[tex]a=\frac{25}{5}=5m/s^2[/tex]

[tex]v=u+at[/tex]

Final velocity when t=2 s

[tex]v=3+5(2)=3+10=13m/s[/tex]

Change in velocity=Final velocity-initial velocity

Using the formula

[tex]\Delta v=13-3=10m/s[/tex]

Hence, the change in velocity=10 m/s

(b) t=0 s and t=2 s

We have to find the displacement ([tex]\Delta x)[/tex]

[tex]\Delta x=ut+\frac{1}{2}at^2[/tex]

Using the formula and substitute t=2

[tex]\Delta r=3(2)+\frac{1}{2}(5)(2)^2=6+10=16 m[/tex]

Hence,the displacement of the particle between t=0 s and t=2 s=16 m