# Mechanics- general motion in a straight line

• MHB
• Shah 72
In summary, the conversation discusses the continuity of two functions, s(t) and v(t), and the values of different variables at specific intervals. There is also a question about finding the value of x, and a mention of a sudden drop in speed at t=4.
Shah 72
MHB

I don't know how to solve this

When are you going to start posting images that are reader friendly?

Both functions have to be continuous ...

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

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

skeeter said:
When are you going to start posting images that are reader friendly?

View attachment 11175
Iam so sorry. I thought it was OK. I will keep it in mind next time

skeeter said:
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 said:
Thank you so much!
I did q(a) as s is continues at t=4
80=2A+4B
Therefore A+2B= 40
skeeter said:
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

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 …

skeeter said:
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!

## 1. What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of how fast an object is moving in a specific direction. In other words, speed is a scalar quantity, while velocity is a vector quantity.

## 2. How is acceleration related to velocity?

Acceleration is the rate of change of velocity over time. This means that an object's velocity will change by a certain amount over a certain period of time, resulting in acceleration. It can be calculated by dividing the change in velocity by the change in time.

## 3. What is the equation for calculating displacement?

The equation for calculating displacement is: displacement = final position - initial position. This means that the displacement is the difference between an object's final position and its initial position in a straight line.

## 4. How does mass affect an object's motion in a straight line?

Mass is a measure of an object's resistance to change in motion. The greater the mass of an object, the more force is needed to accelerate it. This means that objects with a greater mass will have a slower acceleration compared to objects with a smaller mass.

## 5. What is the difference between distance and displacement?

Distance is a measure of how far an object has traveled, while displacement is a measure of the change in an object's position. This means that distance is a scalar quantity, while displacement is a vector quantity. Distance can be greater than or equal to displacement, but it cannot be less than displacement.

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