A rocket sled having an initial speed of 150 mi/hr is slowed by a channel of water. Assume that, during the breaking process, the acceleration is given by a (t) = -kv^2, where v is the velocity and k is a positive constant.

a. Find the velocity and position functions at time t.

b. if it requires a distance of 2000 ft to slow the sled to 15 mi/hr, determine the value of k.

c. Find how many seconds of braking are required to slow the sled to 15 mi/hr.

Step-by-step explanation:

a) since acceleration can be expressed as a (t) = dv/dt

We use the chosen rule to obtain

dv/dt = dv/dx * dx/dt = v * dv/dx

We can now set v dv/dx = - kv^2

Taking v from both sides

dv/dx = - kv

Dv/v = - kdx

In (v) = kx + c

V = e^kx+c

From v (o) = 150

V = 150e^kx

When distance x = 2000ft convert to miles gives 0.378788 mi then velocity v = 15 mi/he

15 = 150. e ^ 0.378788k

Lm (o. t) = - 0.378788k

K = 6.0788mi

C) we return to dv/dt = - kv^2 = dv/v^2 = - k dt

-1/v = - kt + c

From v (o) = 150, we get - 1/150 = c and

1/v - 1/150 = kt

We know v = 15mi/hr and k = 6.0788

1/15 - 1/150 = 6.0788t

T = 9.8704.10^-3hr.