# Velocity calculus

• Sep 10th 2008, 05:55 PM
dataspot
Velocity calculus
The acceleration of a bus is given by $\displaystyle a_{x}(t)= \alpha t$, where $\displaystyle \alpha = 1.27 m/s^3$ is a constant.

If the bus's velocity at time $\displaystyle t_1 = 1.07 s$ is $\displaystyle 5.05 m/s$, what is its velocity at time$\displaystyle t_2 = 2.19 s$?

If the bus's position at time $\displaystyle t_1 = 1.07 s$ is $\displaystyle 5.96 m$, what is its position at time $\displaystyle t_2 = 2.19 s$?

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I tried doing $\displaystyle 5.05m/s + (1.27m/s^3)(2.19s)$, but that can't be right as $\displaystyle (1.27m/s^3)(2.19s)$ is the acceleration of the bus. So I'm not sure how to proceed now.
• Sep 10th 2008, 06:46 PM
dataspot
I've got the first part: the velocity at $\displaystyle t_2$ equals $\displaystyle \int_{1.07}^{2.17} \alpha tdt + 5.05$ which is $\displaystyle 7.37 m/s$.

Still working on the second part.
• Sep 10th 2008, 07:38 PM
dataspot
Still no luck with the second part...
• Sep 10th 2008, 08:11 PM
Shyam
Quote:

Originally Posted by dataspot
The acceleration of a bus is given by $\displaystyle a_{x}(t)= \alpha t$, where $\displaystyle \alpha = 1.27 m/s^3$ is a constant.

If the bus's velocity at time $\displaystyle t_1 = 1.07 s$ is $\displaystyle 5.05 m/s$, what is its velocity at time$\displaystyle t_2 = 2.19 s$?

If the bus's position at time $\displaystyle t_1 = 1.07 s$ is $\displaystyle 5.96 m$, what is its position at time $\displaystyle t_2 = 2.19 s$?

----

I tried doing $\displaystyle 5.05m/s + (1.27m/s^3)(2.19s)$, but that can't be right as $\displaystyle (1.27m/s^3)(2.19s)$ is the acceleration of the bus. So I'm not sure how to proceed now.

Acceleration, $\displaystyle a_x(t)=\alpha t=1.27t$

Velocity: $\displaystyle v(t) = \int a_x(t)~dt= \int 1.27t~dt=\frac{1.27t^2}{2}+c$

$\displaystyle v(t) = 0.635 t^2 +c$

$\displaystyle At\; t_1=1.07, \; v(t_1)=5.05$

$\displaystyle \Rightarrow 5.05 = 0.635 t^2 +c$

$\displaystyle \Rightarrow c=4.32$

$\displaystyle \Rightarrow v(t) = 0.635 t^2 +4.32$

$\displaystyle At \;t_2=2.19, \;v(t_2)= 0.632(2.19)^2 +4.32 = 7.36 \;m/s.$

Position: $\displaystyle x(t) = \int v(t)~dt= \int (0.635 t^2 +4.32)~dt$

$\displaystyle x(t) = \frac{0.635t^3}{3}+0.432t+c$

$\displaystyle x(t) = 0.212t^3+0.432t+c$

$\displaystyle At \; t_1=1.07, \;x(t_1)=5.96$

$\displaystyle \Rightarrow \; 5.96=0.212(1.07)^3+4.32(1.07)+c$

$\displaystyle \Rightarrow \;c=0.11$

$\displaystyle x(t) = 0.212t^3+0.432t+0.11$

$\displaystyle At \; t_2= 2.19,$

$\displaystyle x(t) = 0.212(2.19)^3+0.432(2.19)+0.11$

$\displaystyle x(t) = 10.59 \;m$
• Sep 10th 2008, 08:25 PM
dataspot