Mechanics- general motion in a straight line

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.
  • #1
Shah 72
MHB
274
0
20210604_200321.jpg

I don't know how to solve this
 
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  • #2
When are you going to start posting images that are reader friendly?

motion_prob10.jpg
 
  • #3
Both functions have to be continuous ...

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

$v(t)=\left\{\begin{matrix}
10t & t \in [0,4]\\
\frac{A}{2\sqrt{t}} +B& t \in (4,25]\\
C & t \in (25,50]
\end{matrix}\right.$
 
  • #4
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
 
  • #5
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!
 
  • #6
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
 
  • #7
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 …
 
  • #8
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!
 

FAQ: Mechanics- general motion in a straight line

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.

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.

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.

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.

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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