# PhysicsMechanics- general motion in a straight line

#### skeeter

##### Well-known member
MHB Math Helper
• topsquark

#### skeeter

##### Well-known member
MHB Math Helper
Both functions have to be continuous ...

$s(t)=\left\{\begin{matrix} 5t^2 &t\in [0,4] \\ A\sqrt{t}+Bt & t\in (4,25]\\ Ct+30 & t \in (25,50] \end{matrix}\right.$

$v(t)=\left\{\begin{matrix} 10t & t \in [0,4]\\ \frac{A}{2\sqrt{t}} +B& t \in (4,25]\\ C & t \in (25,50] \end{matrix}\right.$

• Shah 72

#### Shah 72

##### Member
Both functions have to be continuous ...

$s(t)=\left\{\begin{matrix} 5t^2 &t\in [0,4] \\ A\sqrt{t}+Bt & t\in (4,25]\\ Ct+30 & t \in (25,50] \end{matrix}\right.$

$v(t)=\left\{\begin{matrix} 10t & t \in [0,4]\\ \frac{A}{2\sqrt{t}} +B& t \in (4,25]\\ C & t \in (25,50] \end{matrix}\right.$
Thank you so much!!

#### Shah 72

##### Member
Thank you so much!!
I did q(a) as s is continues at t=4
80=2A+4B
Therefore A+2B= 40
Both functions have to be continuous ...

$s(t)=\left\{\begin{matrix} 5t^2 &t\in [0,4] \\ A\sqrt{t}+Bt & t\in (4,25]\\ Ct+30 & t \in (25,50] \end{matrix}\right.$

$v(t)=\left\{\begin{matrix} 10t & t \in [0,4]\\ \frac{A}{2\sqrt{t}} +B& t \in (4,25]\\ C & t \in (25,50] \end{matrix}\right.$
Iam not getting the ans for q(d) Find the value of x

#### skeeter

##### Well-known member
MHB Math Helper
I wasn’t able to find it either …

The sudden “drop” in speed at t = 4 makes the speed function discontinuous there.

$s’(t) = \dfrac{A}{2\sqrt{t}} + B$ becomes $s’(t) = \dfrac{A}{2\sqrt{t}} + B - x$

now, does that make $s(t) = (C-x)t + 30 \text{ and } s’(t) = C-x$ for $t \in (25,50]$ ?

Maybe I’m missing something …

• Shah 72

#### Shah 72

##### Member
I wasn’t able to find it either …

The sudden “drop” in speed at t = 4 makes the speed function discontinuous there.

$s’(t) = \dfrac{A}{2\sqrt{t}} + B$ becomes $s’(t) = \dfrac{A}{2\sqrt{t}} + B - x$

now, does that make $s(t) = (C-x)t + 30 \text{ and } s’(t) = C-x$ for $t \in (25,50]$ ?

Maybe I’m missing something …
Thank you!