2 years ago

The position-time equation for a certain train is xf=2.8m+(8.1m/s)t+(2.9m/s^2)t^2.

1 answer:

3 0

The formula without numbers is
$xf = xi + vxi(t) + \frac{1}{2} (a) {t}^{2}$
do the initial velocity would be
8.1 m/s
and the acceleration would be
2.9 m/s^2

You might be interested in

Question 5. Our results support the idea that if left to freely oscillate, a system will vibrate at a natural frequency that dep

Gekata [30.6K]

Answer:

(b) In ideal condition we neglect mass of spring but in real springs mass of spring adds another factor to its time period.

since we are adding a factor of mass to the system, and frequency being inversely proportional to squared root of mass, we can come to a general conclusion that it effectively reduces the natural frequency .

Explanation:

kindly check the attachment for explanation.