Answer:
Part A) v(t) = u + a t m/sec
Part B) s(t) = s₀ + ut + 0.5 at² meters
Step-by-step explanation:
Given: An object has velocity u m/s at time 0 and constant acceleration a.
Part A) find the velocity after t seconds.
We should know that: [tex]a = \frac{dv}{dt}[/tex]
By integrating both sides with respect to the time t
∴∫ dv = ∫ a dt
∴ v(t) = a t + constant
the object has velocity u m/s at time 0 ⇒ constant = u
∴ v(t) = u + a t m/sec
Part B) find the displacement of the object between time 0 and time t.
We should know that: [tex]v = \frac{ds}{dt}[/tex]
By integrating both sides with respect to the time t
∴∫ ds = ∫ v dt
∵ v(t) = u + a t
∴ ∫ ds = ∫ (u + a t) dt
∴ s(t) = ut + 0.5 at² + constant
Let at t = 0 displacement = s₀ ⇒ ∴ Constant = s₀
∴ s(t) = s₀ + ut + 0.5 at² meters
a. The expression of velocity is, [tex]v(t)=u+at[/tex]
b. The expression of displacement is, [tex]s(t)=ut+\frac{1}{2}at^{2} [/tex]
Acceleration is defined as the rate of change of velocity with respect to time.
[tex]a=\frac{dv}{dt} \\ \\ dv=adt[/tex]
Integrate both side.
[tex]v=at+c[/tex]
Given that, at time t = 0, v = u
Substitute ,
[tex]u+a*0=c\\ \\ c=u[/tex]
So that expression of velocity is,
[tex]v(t)=u+at[/tex]
Velocity is defined as the rate of change of displacement with respect to time.
[tex]v(t)=\frac{ds}{dt}\\ \\ ds=v(t)dt[/tex]
Integrate on both side.
[tex]ds=(u+at)dt\\ \\ s(t)=ut+\frac{1}{2}at^{2} [/tex]
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