Couple questions on acceleration and velocity

[pucks214]

1. Santa loses his footing and slides down a frictionless, snowy roof that is tilted at an angle of 35°. If Santa slides 6 m before reaching the edge, what is his speed as he leaves the roof?

2. 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 18 s, then the motor stops. The rocket altitude 22 s after launch is 4900 m. You can ignore any effects of air resistance.
(a) What was the rocket's acceleration during the first 18 s? m/s2
(b) What is the rocket's speed as it passes through a cloud 4900 m above the ground? m/s

3. You are driving to the grocery store at 22 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
a. What acceleration (magnitude) will bring you to rest right at the intersection?
b. How long (total) does it take you to stop?

4. A speed skater moving across frictionless ice at 8.8 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s. What is her acceleration on the rough ice?

Equations given: x = Vo(t) + .5at^2
Vf = Vo +at
Vf^2 = Vo^2 +2a(delta x)

Note: may also need to use vectors.

Thank you.

 

[rl.bhat]

Hi pucks214,
Welcome to PF.
Before asking for help you have to show your attempt.

 

[tomcue]

1. Santa's speed as he leaves the roof can be calculated using the equation Vf^2 = Vo^2 + 2a(delta x), where Vf is the final velocity, Vo is the initial velocity (in this case, 0 m/s), a is the acceleration, and delta x is the distance traveled. Plugging in the given values, we get Vf^2 = 0^2 + 2(9.8 m/s^2)(6 m), which simplifies to Vf = 11.2 m/s. Therefore, Santa's speed as he leaves the roof is 11.2 m/s.

2. (a) The rocket's acceleration during the first 18 s can be calculated using the equation Vf = Vo + at, where Vf is the final velocity (in this case, 0 m/s), Vo is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get 0 = Vo + a(18 s), which simplifies to a = -Vo/18. Since the rocket is launched straight up, Vo is equal to the initial velocity of the rocket, which is 0 m/s. Therefore, the rocket's acceleration during the first 18 s is 0 m/s^2.

(b) The rocket's speed as it passes through a cloud 4900 m above the ground can be calculated using the equation Vf^2 = Vo^2 + 2a(delta x), where Vf is the final velocity, Vo is the initial velocity, a is the acceleration, and delta x is the distance traveled. Plugging in the given values, we get Vf^2 = 0^2 + 2(-9.8 m/s^2)(4900 m), which simplifies to Vf = 99.0 m/s. Therefore, the rocket's speed as it passes through the cloud is 99.0 m/s.

3. (a) The acceleration needed to bring the car to rest at the intersection can be calculated using the equation x = Vo(t) + .5at^2, where x is the distance traveled, Vo is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get 110 m = (22 m/s)(0.50 s) + .5a(0.50 s)^2,