Motion problem - Shock drop ride

[09jk]

Homework Statement

a ride at a them park is a vertically ascending/decending ride. The total mass of the 10 person carriage is 750kg (people included). the ride rises up to the top of the ride until stationary waits for 2 seconds and then drops at a rate of 10ms^-2 for 3.5m. when near the bottom of the ride, the carriages brakes are applied for 0.8m.

a) how long is the carriage in freefall until the brakes are applied?

Homework Equations

not too sure, but i think you'll probably require:
v=u+at
s=1/2(u+v)t

The Attempt at a Solution

to be honest, i have no idea where to begin
thanks

 

[skand]

hello,I suppose you'ld not need most of the data given.
We can assume $$g=10ms^{-2}$$

now we know distance covered,the acceleration,the initial velocity.
Just put in formulas

 

[nlightner]

for any help

I would approach this problem by first calculating the initial velocity of the carriage as it starts its descent. We can use the equation v = u + at, where v is the final velocity (0 m/s since it comes to a stop), u is the initial velocity (unknown), a is the acceleration (10 ms^-2), and t is the time (2 seconds). Solving for u, we get u = 20 m/s.

Next, we can use the equation s = 1/2(u+v)t to find the distance the carriage travels during the 2 seconds it is stationary at the top. We know s (the distance) is 0 since it is stationary, u (the initial velocity) is 20 m/s, v (the final velocity) is 0 m/s, and t (the time) is 2 seconds. Solving for s, we get s = 20 meters.

Now, we can use the equation s = ut + 1/2at^2 to find the distance the carriage travels during the 3.5 meters of freefall. We know s (the distance) is 3.5 meters, u (the initial velocity) is 20 m/s, a (the acceleration) is 10 ms^-2, and t (the time) is unknown. Solving for t, we get t = 0.5 seconds.

Therefore, the carriage is in freefall for 0.5 seconds before the brakes are applied.