(a) At the moment `t=0`, a particle leaves the origin and moves in the positive direction of the x-axis. Its velocity is given by `v=10-2t`. Find the displacement and distance in the first `8s`.
(b) If `v=t-t^(2)`, find the displacement and distamce in first 2s.

Explanation

(a) `v=10-2t`
`(ds)/(dt)=10-2t`
`int_(0)^(s) ds=int_(0)^(8)(10-2t) dt`
`|s|_(0)^(s)=10t-2t^(2)/2=|10t-t^(2)|_(0)^(8)`
`s=10(8)-(8)^(2)=16 m`
Displacement in the first 8 s is `16 m`.
To calculate the distance, first check whether the velocity changes sign time interval.
`v=0=10-2timpliest=5 s`
At` t=5 s`, the velocity changes sign
`a=(dv)/(dt)=-2`
`0 lt t lt 5 s, v gt 0, a lt 0`, the particle is moving along the positive x-axis and slowing down.
`5 s lt t lt = 8 s, v lt 0, a lt 0`, the particle is moving along the negative x-axis and is speeding up.
Distance`=int_(0)^(5) v dt+|int_(5)^(8) v dt|`
Take this value as positive
`=int_(0)^(5)(10-2t)dt+|int_(5)^(8)(10-2t) dt|`
`=|10t-t^(2)|_(0)^(5)+||10t-t^(2)|_(5)^(8)|`
`=(50-25)+|(10xx8-8^(2))-(10xx5-5^(2))|`
`=25+|16-25|`
`=25+|-9|`
`=25+9=34 m`
Note: `|x|=x,|-x|=x,|-2|=2`.
(b) `v=t-t^(2)`
` (ds)/(dt)=t-t^(2)`
`int_(0)^(s) ds =int_(0)^(2)(t-t^(2)) dt`
`|s|_(0)^(s)=|t^(2)/2-t^(3)/3|_(0)^(2)`
`s=((2))^(2)/2-((2))^(3)/3=2-8/3=-2/3 m`
Displacement in the first `2s =-2/3 m`
To calculate the distance, check whether the velocity changes sign in the time interval.
`v=t-t^(2)=t(1-t)=0 impliest=0, t=1 s`
At t=0, the particle is at the origin.
At `t=1 s`, the velocity changes sign.
`v=t-t^(2)`
`a=(dv)/(dt)=1-2t`
`a=0=1-2timpliest=1/2 s`
`0 lt t lt 1/2 s, v gt 0 ,a gt 0`, the particle is moving along the positive x-axis and is speeding up.
`1/2 s lt t lt 1 s, v gt 0, a lt 0`, the particle is moving along the positive x-axis and is slowing down.
`1s lt t lt = 2 s, v lt 0, a lt 0`, the particle is moving along the negative x-axis is speeding up.
Distance`=int_(0)^(1) v dt+|int_(1)^(2) v dt| `
Take this value as positive
`=|t^(2)/2-t^(3)/3|_(0)^(1)+||t^(2)/2-t^(3)/3|_(1)^(2)|`
`=(1/2-1/3)-(0)+|{(2^(2)/2-2^(3)/3)-(1^(2)/2-1^(3)/3)}|`
`=1/6+|-2/3-1/6|=1/6+|-5/6|`
`=1/6+5/6=1 m`