The equation of motion of a particle is s= 2y^3 - 7t^2 + 4t +1 where s is in meters and t is in seconds
a) Find the velocity and acceleration as functions of t.
b) Find the acceleration after 1 second.
are you sure that is 2y^3 and not 2 t^3 ?
2 y^3 is all right, but unusual and probably not intended.
If as typed, the 2y^3 has nothing to do with the velocity and acceleration
v =ds/dt = -14 t + 4
a = dv/dt = -14
If as I kind of have a hunch you mean
v = 6 t^2 - 14 t + 4
a = 12 t - 14
it was 2 t^3
Thanks
To find the velocity and acceleration as functions of time (t), we need to take the derivatives of the equation of motion with respect to time.
a) Velocity (v):
To find the velocity, we need to take the derivative of the equation of motion with respect to time (t). Let's differentiate the equation:
s = 2y^3 - 7t^2 + 4t + 1
Differentiating both sides with respect to time:
ds/dt = d/dt(2y^3) - d/dt(7t^2) + d/dt(4t) + d/dt(1)
The derivative of y^3 with respect to t is 0 since y does not appear in the equation. Also, the derivative of a constant term (1) with respect to t is 0. Therefore, we can simplify the equation:
ds/dt = -14t + 4
So, the velocity (v) as a function of time is:
v = ds/dt = -14t + 4
b) Acceleration (a):
To find the acceleration, we need to take the derivative of the velocity with respect to time. Let's differentiate the equation for velocity:
v = -14t + 4
Differentiating both sides with respect to time:
dv/dt = d/dt(-14t) + d/dt(4)
The derivative of the constant term 4 with respect to t is 0. Therefore, we can simplify the equation:
dv/dt = -14
So, the acceleration (a) as a function of time is:
a = dv/dt = -14
To find the acceleration after 1 second (t = 1), we substitute t = 1 into the equation for acceleration:
a = -14
Therefore, the acceleration after 1 second is -14 m/s^2.