Respuesta :
In order to answer this excercise you need to use the formulas
s = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
s = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Margy starts at 1.4 m/s
and accelerates 0.2 m/s^2 over a distance of 100.0m. The velocity over this time period will look like
v(t) = 1.4 + 0.2 *t which tells you the velocity with respect to the time
the distance over this time. Another formula can also be used for this:
x(t) = Xi+Vi*t + 1/2at^2
Xi is the initial position taken to be 0 here
Vi is initial speed which is 1.4 m/s a is 0.2 m/s^2
Total distance was 100
100= 0+1.4*t + 1/2 * 0.2 * t^2
Solve for t and then use the fact that v=1.4+0.2*t, to find the final velocity.
x(t) = Xi+Vi*t + 1/2at^2
Xi is the initial position taken to be 0 here
Vi is initial speed which is 1.4 m/s a is 0.2 m/s^2
Total distance was 100
100= 0+1.4*t + 1/2 * 0.2 * t^2
Solve for t and then use the fact that v=1.4+0.2*t, to find the final velocity.