kloseyq

2022-01-12

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation $x\left(t\right)=\alpha {t}^{2}-\beta {t}^{3}$, where and
Calculate the average velocity of the car for each time interval:
a) $t=0$ to $t=2.00s$
b) $t=0$ to
c) $t=2.00s$ to $t=4.00s$

chumants6g

Expert

Step 1
First we have to calculate the distance x at each t.

$x\left(2\right)=1.50{\left(2\right)}^{2}-0.0500{\left(2\right)}^{3}=5.6m$
$x\left(4\right)=1.50{\left(4\right)}^{2}-0.0500{\left(4\right)}^{3}=20.8m$
The average velocity is the displacement divided by the time interval at which this displacement happened.

$\mathrm{\Delta }x={x}_{f}-{x}_{i}$
$\mathrm{\Delta }t={t}_{f}-{t}_{i}$
Step 2
a)

$x\left(2\right)=1.50{\left(2\right)}^{2}-0.0500{\left(2\right)}^{3}=5.6m$

$=2.8ms$
Step 3
b)
$x\left(4\right)=1.50{\left(4\right)}^{2}-0.0500{\left(4\right)}^{3}=20.8m$

$=5.2ms$
Step 4
c) $x\left(2\right)=1.50{\left(2\right)}^{2}-0.0500{\left(2\right)}^{3}=5.6m$
$x\left(4\right)=1.50{\left(4\right)}^{2}-0.0500{\left(4\right)}^{3}=20.8m$

$=7.6ms$

Mary Herrera

Expert

Step 1
a) The equation for cars

nick1337

Expert

Step 1
a) The position of the car as a function of time t is given by
$x\left(t\right)=\alpha {t}^{2}-\beta {t}^{3}$
where
$\alpha =1.50m/{s}^{2}\phantom{\rule{0ex}{0ex}}\beta =0.05m/{s}^{3}$
The average velocity is given by the ratio between the displacement and the time taken:
$v=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$
The position at t = 0 is:
$x\left(0\right)=\alpha ×{0}^{2}-\beta ×{0}^{3}=0$
The position at t = 2.00 s is:
$x\left(2\right)=\alpha ×{2}^{2}-\beta ×{2}^{3}=5.6m$
So the displacement is
$\mathrm{\Delta }x=x\left(2\right)-x\left(0\right)=5.6-0=5.6m$
The time interval is
$\mathrm{\Delta }t=2.0s-0s=2.0s$
And so, the average velocity in this interval is
$v=\frac{5.6m}{2.0s}=2.8m/s$
b) The position at t = 0 is:
$x\left(0\right)=\alpha ×{0}^{2}-\beta ×{0}^{3}=0$
While the position at t = 4.00 s is:
$x\left(4\right)=\alpha ×{4}^{2}-\beta ×{4}^{3}=20.8m$
So the displacement is
$\mathrm{\Delta }x=x\left(4\right)-x\left(0\right)=20.8-0=20.8m$
The time interval is
$\mathrm{\Delta }t=4.00-0=4.0s$
So the average velocity here is
$v=\frac{20.8}{4.0}=5.2m/s$
c) The position at t = 2 s is:
$x\left(2\right)=\alpha ×{2}^{2}-\beta ×{2}^{3}=5.6m$
While the position at t = 4 s is:

$x\left(4\right)=\alpha ×{4}^{2}-\beta ×{4}^{3}=20.8m$
So the displacement is
$\mathrm{\Delta }x=20.8-5.6=15.2m$
While the time interval is
$\mathrm{\Delta }t=4.0-2.0=2.0s$
So the average velocity is
$v=\frac{15.2}{2.0}=7.6m/s$

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