The automobile is originally at rest at s = 0. If its speed is increased by dv/dt = (0.05t2)ft/s2, where t is in seconds, determine magnitudes of its velocity and acceleration when t = 18 s.
V=dv/dt * t + vi
acceration is constant, a=dv/dt
Notice acceleration is not constant, but you can figure it from dv/dt at the given time.
smd wei
To determine the magnitude of velocity and acceleration of the automobile at t = 18 s, we need to perform integration twice.
First, let's find the velocity of the automobile. Given the rate of change of velocity, dv/dt = 0.05t^2 ft/s^2, we need to integrate this expression with respect to time (t) to obtain the expression for velocity (v).
∫dv/dt dt = ∫(0.05t^2) dt
Integrating the above expression gives:
v = ∫(0.05t^2) dt = 0.05∫t^2 dt = 0.05(t^3/3) + C1
Here, C1 is the constant of integration.
Now, let's find the acceleration of the automobile. We know that acceleration is the derivative of velocity with respect to time. So, we need to differentiate the expression for velocity (v) with respect to time (t).
d/dt (0.05(t^3/3) + C1) = 0.05(t^2)
Now, we have the expression for acceleration as a function of time.
To determine the magnitudes at t = 18 s, substitute t = 18 into the expressions for velocity and acceleration:
v = 0.05(18^3/3) + C1
a = 0.05(18^2)
Evaluate these expressions using a calculator to find the numerical values of velocity and acceleration.