Answer:
Explanation:
Given
acceleration of a particle is given by [tex]a(t)=pt^2-qt^3[/tex]
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=pt^2-qt^3[/tex]
[tex]dv=(pt^2-qt^3) dt[/tex]
Integrating we get
[tex]\int dv=\int \left ( pt^2-qt^3\right )dt[/tex]
[tex]v=\frac{pt^3}{3}-\frac{qt^4}{4}+c_1[/tex]
at [tex]t=0\ v=0[/tex]
therefore [tex]c_1=0[/tex]
We know velocity is given by
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]vdt=dx[/tex]
integrating
[tex]\int dx=\int \left ( \frac{pt^3}{3}-\frac{qt^4}{4}+\right )dt[/tex]
[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}+c_2[/tex]
using conditions
at [tex]t=0\ x=0[/tex]
[tex]c_2=0[/tex]
[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}[/tex]
(a) The velocity of the particle as function of time is pt³/3 - qt⁴/4 + C.
(b) The position of the particle as function of time is pt⁴/12 - qt⁵/20 + Ct + C.
The velocity of the particle is the integral of the acceleration of the particle.
The velocity of the particle is calculated as follows;
v = ∫a(t)
v = ∫(pt² - qt³)dt
v = pt³/3 - qt⁴/4 + C
The position of the particle as function of time is the integral of velocity of the particle.
x = ∫v
x = ∫(pt³/3 - qt⁴/4 + C)dt
x = pt⁴/12 - qt⁵/20 + Ct + C
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