A particle traveling in a strainght line passes a fixed point O

[karush]

View attachment 5141
so if $a\left(t\right)=p+qt$

then $v\left(t\right)=\int a\left(t\right) dt = pt+\frac{q {t}^{2}}{2}+C$

if $v=3.5$ when $t=2$ then $1.75=p+q$

if so, now what? the answers are $4, -3; 4m$

 

[MarkFL]

You can get a second equation involving the parameters $p$ and $q$ by using the information regarding when the particle comes to rest...:)

 

[karush]

$v\left(t\right)=\frac{q{t}^{2}}{2}+pt+1.5$

$v\left(2\right)=2p+2p+1.5$

$v\left(3\right)=3p+\frac{9q}{2}+1,5$

Solving simultaneously
$3.5=v\left(2\right)$
$0=v\left(3\right)$
$p=4\ q=-3$

$v1\left(t\right)=\frac{-3{t}^{2}}{2}+4t+1.5$

$\int_1 ^2 v1\left(t\right)dt=4$

 

[MarkFL]

I would first determine the velocity function:

\(\displaystyle a(t)=\d{v}{t}=qt+p\)

Integrate:

\(\displaystyle \int_{\frac{3}{2}}^{v(t)}\,du=\int_0^t qw+p\,dw\)

\(\displaystyle v(t)=\frac{q}{2}t^2+pt+\frac{3}{2}\)

Now, use the two data points to determine the parameters:

\(\displaystyle v(2)=\frac{q}{2}2^2+p(2)+\frac{3}{2}=2q+2p+\frac{3}{2}=\frac{7}{2}\implies p+q=1\)

\(\displaystyle v(3)=\frac{q}{2}3^2+p(3)+\frac{3}{2}=\frac{9}{2}q+3p+\frac{3}{2}=0\implies 2p+3q=-1\)

Solving this system, we obtain:

\(\displaystyle (p,q)=(4,-3)\)

Hence, the velocity is:

\(\displaystyle v(t)=-\frac{3}{2}t^2+4t+\frac{3}{2}\)

And so we may state:

\(\displaystyle \d{x}{t}=-\frac{3}{2}t^2+4t+\frac{3}{2}\)

Integrate:

\(\displaystyle \int_0^{x(t)} \,du=\int_0^t -\frac{3}{2}w^2+4w+\frac{3}{2}\,dw\)

\(\displaystyle x(t)=-\frac{1}{2}t^3+2t^2+\frac{3}{2}t\)

And thus:

\(\displaystyle |x(2)-x(1)|=\left|-\frac{1}{2}2^3+2(2)^2+\frac{3}{2}(2)+\frac{1}{2}1^3-2(1)^2-\frac{3}{2}(1)\right|=\left|-4+8+3+\frac{1}{2}-2-\frac{3}{2}\right|=\left|4\right|=4\)