Velocity calculus

**dataspot** · September 10th 2008, 05:55 PM

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

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

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

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

**dataspot** · September 10th 2008, 06:46 PM

I've got the first part: the velocity at $t_2$ equals $\int_{1.07}^{2.17} \alpha tdt + 5.05$ which is $7.37 m/s$ .

Still working on the second part.

**dataspot** · September 10th 2008, 07:38 PM

Still no luck with the second part...

**Shyam** · September 10th 2008, 08:11 PM

Originally Posted by dataspot

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

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

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

----

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

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

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

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

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

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

$\Rightarrow c=4.32$

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

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

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

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

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

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

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

$\Rightarrow \;c=0.11$

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

$<br /> At \; t_2= 2.19,$

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

$x(t) = 10.59 \;m$

**dataspot** · September 10th 2008, 08:25 PM

Thanks, but I don't really follow your math at all.

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