The position-time equation for a certain train is xf=2.2m+(8.6m/s)t+(2.3m/s2)t2. what is the position of the train in the previous problem at the time 4.9 s ?

The position of the train at the time 4.9 seconds is 99.563 meters

Explanation:

* Lets explain how to solve the problem

The position-time equation for a certain train is:

f(x) = 2.2 + 8.6 t + 2.3 t², where f(x) is the position of the train in

meters at a certain time t in second.

We need to find the position of the train at the time 9.4 seconds.

So lets substitute the value of t in the position-time equation.

The position-time equation f(x) = 2.2 + 8.6 t + 2.3 t²

The time = 4.9 seconds

f(x) = 2.2 + 8.6(4.9) + 2.3(4.9)²

f(x) = 2.2 + 42.14 + 55.223

f(x) = 99.563

The position of the train at the time 4.9 seconds is 99.563 meters

c. magnitude and direction

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg