A particle is moving along a straight line and the acceleration is given by a(t)=3t+5, (units m/s2). a)Find v(t) , the velocity at time t if the initial velocity is v(0) = -4 ft/s? b)Find the total distance traveled during the time interval [0, 2]?

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aristeus aristeus · Answer 1 · 2019-09-25T09:03:21+02:00

Answer:

Velocity will be [tex]v(t)=\frac{3t^2}{2}+5t-4ft/sec[/tex]

Distance traveled = 6 feet

Explanation:

We have given acceleration [tex]a(t)=3t+5[/tex]

We know that [tex]v(t)=\int a(t)dt[/tex]

So [tex]v(t)=\int 3t+5dt[/tex]

[tex]v(t)=\frac{3t^2}{2}+5t+c[/tex]

We have given [tex]v(0)=-4ft/sec[/tex]

So [tex]-4=\frac{3\times 0^2}{2}+5\times 0+c[/tex]

c = -4 ft/sec

So [tex]v(t)=\frac{3t^2}{2}+5t-4ft/sec[/tex]

Now we have to find distance traveled in interval [0,2]

So distance [tex]s=\int v(t)dt[/tex]

[tex]s=\int_{0}^{2}v(t)dt[/tex]

[tex]s=\int_{0}^{2}(\frac{3t^2}{2}+5t-4)dt[/tex]

[tex]s=(\frac{t^3}{2}+\frac{5t^2}{2}-4t)_{0}^{2}[/tex]

[tex]=(4+10-8)-0[/tex]

[tex]=6ft[/tex]