Net displacement of a particle given its equation of motion

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javii
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Homework Statement


Hello PF,

I need some help with the assignment given:
v=18-2t^2 m/s, where t is in second. When t= 0 the position of the particle is s_0 = - 3 m.
For the first 5 seconds.
Determine the total distance ( i got it to 65,33 ft) and the net displacement Δs, and the value of s at the end of the interval.

The Attempt at a Solution


I guess I have to use the formula:
s(t)=s_0 + ∫_0 ^t v(t) dt

s(t)=3+∫_0^3 18-2t^2 dt
 
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javii said:
s(t)=s_0 + ∫_0 ^t v(t) dt
That is the correct equation.
S0= -3m.
 
Be careful with signs and units and integral borders. Apart from that, the total distance is fine, and the formula for s(t) will help with the net displacement Δs and s at the end.
 
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cnh1995 said:
That is the correct equation.
S0= -3m.
But when I integrate it, I do not get the correct answer.
 
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javii said:
But when I integrate it, I do not get the correct answer.
Why are you multiplying the integral by -3?
Also, the upper limit of the integral is not 3.
gneill said:
... and tf = 5 seconds.
 
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cnh1995 said:
Why are you multiplying the integral by -3?
Also, the upper limit of the integral is not 3.
May bad. So this should be correct?
 
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