I think it is 500 cm. Hope I helped!
Answer:
<em>The boat takes 2.5 seconds to slow down to 5 m/s</em>
Explanation:
<u>Motion With Constant Acceleration</u>
It's a type of motion in which the velocity of an object changes uniformly in time. The equation that rules the change of velocities is:
![v_f=v_o+at\qquad\qquad [1]](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat%5Cqquad%5Cqquad%20%5B1%5D)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
Using the equation [1] we can solve for a:
The acceleration produced by the friction of water is 
and the boat is initially traveling at v0=30 m/s. When the motor is shut off, the boat will start braking until it stops. We need to find the time it takes to ready the final speed of vf=5 m/s.
Let's solve the above equation for t:
t = 2.5 s
The boat takes 2.5 seconds to slow down to 5 m/s
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
