Respuesta :
Answer:
The displacement of the block is 9 m.
Explanation:
Given that,
Mass of block = 2 kg
Time = 3 sec
The force is
[tex]F(t)=4t[/tex]
The acceleration is
[tex]a=\dfrac{F}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{4t}{2}[/tex]
[tex]a=2t[/tex]
We need to calculate the velocity
Using formula of acceleration
[tex]v=\int_{0}^{t}{a dt}[/tex]
Put the value into the formula
[tex]v=\int_{0}^{3}(2t dt)[/tex]
[tex]v=(\dfrac{2t^2}{2})_{0}^{3}[/tex]
[tex]v=(t^2)_{0}^{3}[/tex]
[tex]v=9\ m/s[/tex]
We need to calculate the displacement
Using formula of velocity
[tex]s=\int_{0}^{3}(v dt)[/tex]
[tex]s=(t^2)_{0}^{3}[/tex]
[tex]s=(\dfrac{t^3}{3})_{0}^{3}[/tex]
[tex]s=9\ m/s[/tex]
Hence, The displacement of the block is 9 m.
Answer:
[tex]s=9\ m[/tex]
Explanation:
Given:
mass of the block,[tex]m=2\ kg[/tex]
force as a function of time, [tex]F(t)=4t[/tex]
time of observation, [tex]t=3\ s[/tex]
initial velocity of the block, [tex]u=0\ m.s^{-1}[/tex]
Now the acceleration:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{4t}{2}[/tex]
[tex]a=2t[/tex]
Now the related velocity:
[tex]v=\int {a} \, dt[/tex]
[tex]v=\int 2t\ dt[/tex]
[tex]v=t^2[/tex]
Integrating again we get the displacement:
[tex]s=\int\limits^3_0 v\ dt[/tex]
[tex]s=\int\limits^3_0 t^2\ dt[/tex]
[tex]s=\frac{t^3}{3}\vert^3_0[/tex]
[tex]s=9\ m[/tex]
