Respuesta :
Answer:
The displacement and distance is 22 m and 20.09 m.
Explanation:
Suppose, a(t)=11
v₀=-1 ; 0<t<2.....these are meant to be less than or equal to symbols
Given that,
Acceleration = 11 m/s²
We need to calculate the displacement
Using equation of velocity
[tex]\dfrac{dv}{dt}=11[/tex]
[tex]dv=11dt[/tex]
On integration
[tex]\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{11dt}[/tex]
[tex]v-v_{0}=11t[/tex]
[tex]v=v_{0}+11t[/tex]
Put the value of v₀
[tex]v=-1+11t[/tex]
[tex]\dfrac{dx}{dt}=-1+11t[/tex]
[tex]dx=(-1+11t)dt[/tex]
[tex]\int_{x_{i}}^{x_{f}}{dx}=\int_{0}^{t}{(-1+11t)dt}[/tex]
[tex]x_{f}-x_{i}=(\dfrac{11t^2}{2})_{0}^{2}[/tex]
[tex]x_{f}-x_{i}=\dfrac{11}{2}\times4[/tex]
[tex]x_{f}-x_{i}=22\ m[/tex]
The displacement is 22 m.
We need to calculate the distance
Using formula of distance
[tex]s=\int_{0}^{2}{|v|dt}[/tex]
[tex]s=\int_{0}^{\frac{1}{11}}{(-1+11t)dt}+\int_{\frac{1}{11}}^{2}{(-1+11t)dt}[/tex]
[tex]s=(t-\dfrac{11t^2}{2})_{0}^{\frac{1}{11}}+(-t+\dfrac{11t^2}{2})_{\frac{1}{11}}^{2}}[/tex]
[tex]s=\dfrac{1}{11}-\dfrac{1}{22}-2+22+\dfrac{1}{11}-\dfrac{1}{22}[/tex]
[tex]s=22.09\ m[/tex]
Hence, The displacement and distance is 22 m and 20.09 m.
The displacement of the particle is 20m
Calculating the displacement:
Given that the acceleration is:
a(t) = 11 m/s²
we know that acceleration:
a(t) = dv/dt
dv = a(t)dt
[tex]\int\limits^v_{v_0} {} \, dv=\int\limits^t_0 {a} \, dt[/tex]
[tex]v-v_0=at[/tex]
where v₀ is the initial velocity = -1 m/s,
so,
v = 11t - 1 m/s
Now,
v = dx/dt
where x is the displacement
dx = vdt
[tex]\int\limits^{}_{} {x} \, dx =\int\limits^2_0 {v} \, dt\\\\x= \int\limits^2_0 ({11t-1}) \, dt\\\\x=[\frac{11t^2}{2}-t]_0^2\\\\[/tex]
x = 20m
Learn more about equations of motion:
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