turtle goes 0.50m/s rabbit goes 3.0m/s^2.. rabbit gives turtle 195 seconds head start. how long will it take for hare to catch turtle?
195*.5=97.5m
+amount covered while rabbit is catching up
d = 0.5m/s * 195s = 97.5 m. Head-start.
d2 = d1 + 195 m.
0.5*a*t^2 = r*t + 195
0.5*3*t^2 = 0.5*t + 195
1.5t^2 - 0.5t - 195 = 0
Use Quadratic formula.
t = 11.6 s.
To find out how long it will take for the hare to catch the turtle, we need to determine how far the turtle will be ahead when the hare starts.
Given that the turtle has a speed of 0.50 m/s and the rabbit has an acceleration of 3.0 m/s^2, we can use the equation for displacement with constant acceleration:
s = ut + (1/2)at^2
Where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration.
For the turtle, the initial velocity is 0.50 m/s and the time is 195 seconds (the head start given to the turtle). Plugging these values into the equation, we can find the displacement of the turtle:
s_turtle = (0.50 m/s)(195 s) + (1/2)(0 s^2)(3.0 m/s^2)
s_turtle = 97.5 m
So the turtle has a head start of 97.5 meters.
Now, we need to calculate the distance the rabbit will cover in the time it takes to catch up with the turtle. Since the rabbit has an acceleration of 3.0 m/s^2, it will constantly increase its speed until it catches the turtle.
Using the equation for displacement with constant acceleration again, and assuming that the hare catches the turtle at time t_catch, we have:
s_rabbit = (0 m/s)(t_catch s) + (1/2)(3.0 m/s^2)(t_catch s)^2
s_rabbit = (3/2) t_catch^2 m
Since the turtle has a head start of 97.5 meters, the distance covered by the rabbit must be equal to the head start for them to meet:
97.5 m = (3/2) t_catch^2 m
Now, we can solve this equation to find t_catch, which is the time it takes for the hare to catch the turtle.