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

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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(t) = αt2 - βt3, where α = 1.50 m/s2 and β = 0.0500 m/s3. Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.00 s; (b) t = 0 to t = 4.00 s; (c) t = 2.00 s to t = 4.00 s.

Physics

1 answer:

serg [3.5K] · Answer 1 · 2026-02-24T04:35:40+03:00

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

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