Kinematics? Velocity and Acceleration problems... please help...

Question

1.) A police car is searching for a fugitive that managed to escape a while ago. Knowing that he is now safe,the fugitive begins to take a rest until he notices a police car approaching him 10m/s, accelerating at 5m/s^2 and it is 100m away. The fugitive grabs a motorcycle and starts it accelerating at the same rate as the police car. How much time will it take the police car to catch the fugitive?

2.) A car starts from rest and moves along a straight line with an acceleration of a=(3s^ -1/3)m/s^2, where s is in meters. Determine the car's velocity and position when t=10s.

thanks a lot...

Answer 1

These sound like dynamics problems, so I'll answer #2 first...

2.) Using V_t = V_o +a*t, where V_o is initial velocity, a = (3^-1/3), and t = 10s, we simply plug into the equation and receive====> 6.933 m/s

NOTE: I'm assuming {a=(3s^ -1/3)m/s^2} was a typo, and should have read (3^-1/3)m/s^2....

1.) OK, dynamics...here we go

Since they meet up at the same distance and same time, use the equations  

(a): d(t) = d(0) + v(0)*t +1/2*a*t^2.....for the 5-0 (police)

=====> (a): rearranged is...delta_d = 10*t +1/2*(5)*t^2

also use equation

(b): v(t)^2 = v(0)^2 + 2*a*(delta_d)......for the motorcycle

=====> (b): v(t) = sqrt(2*a*(delta_d))

We still dont know v(t)^2 for the motorcycle, so use

(c): v(t) = v(0) + a*t

Now, plug v(t) from (c) into (b) and solve for delta_d...

Now, use your delta_d to solve for time using equation (a)...

NOTE: i DONT FEEL LIKE SOLVING ALL THIS OUT, BUT HERE ARE THE STEPS, AND YOU;LL ENJOY THE ANSWER. BONJOUR