To determine
(a)
To find:
The displacement
Answer
1756 meters
Explanation
1) Concept:
i) The displacement of a particle is calculated by using the integrating velocity function that is t=∫vtdt
ii) ∫ab[fx+gx] dx=∫abfxdx+∫abgxdx
iii) ∫xmdx=xm+1m+1+C
2) Given:
vt=t2-t the velocity function measured in meter per second.
3) Calculation:
The given velocity of the particle is vt=t2-t
Therefore, the displacement of the particle can be calculated by using the formula
st=∫vtdt
st=∫(t2-t) dt
st=∫(t2) dt-∫t dt
Integrate using the property
st=t33-t22+C
To find the displacement over [0, 5] integrate the function s(t) over the interval [0, 5]
s=∫05(t2-t) dt
s=∫05(t2) dt-∫05t dt
s=t3305-t2205
Substitute the limits
s=533-0-522-0
s=1253-252
s=1756m
Conclusion:
Therefore, displacement oftheparticle is 1756 meters
To determine
(b)
To find:
The distance travelled by the particle during the time interval [0, 5]
Answer
1766 meters
Explanation
Consider velocity function from part a)
To find the distance travelled by theparticle during the time interval [0, 5] integrate the function |v(t)| in the interval [0, 5]
As the zeros functions (t2-t) are at t=0 and t=1 . We can see that in the interval [0, 1], the function is negative and in [1,4] positive, so consider the function |v(t)| as t-t2 in [0,1].
So to find the distance, split the integration from 0 to 1 and 1 to 5
Distance=∫01(t-t2) dt+ ∫15(t2-t) dt
Integrate these functions
D=t2201-t3301+t3315-t2215
Substitute the limits
D=12-13+ 533-13-522-12
D= 16+ 1253-252
D=16+175 6=1766
Conclusion:
Therefore,
Distance travelled by the particle in time interval [0, 5] is 1766 meters