Please help me on 2 physics problems?

Question

1) After landing, a jetliner on a straight runway taxis to a stop at average velocity of -35.0km/hr. If the plane takes 7.00s to come to rest, what are the plane's initial velocity and acceleration?

Ans: -70.0km/hr or -19.4m/s, +2.78m/s^2

I know I should use V = Vo + at to find a,

but what equation I should use to find the initial velocity??

2) A hockey puck sliding along the ice hits the boards head-on moving to the left with a speed of 35 m/s. As it reverses direction, it is in contact with the boards for 0.095s, before rebounding at slower speed of 11m/s. Determine the average acceleration the puck experienced while hitting the boards. Typical car accelerations are 5m/s^2. Comment the size of your answer, and why it is so different from this value especially when the puck speeds are similar to car speeds.

Ans: 4.8 x 10^2

I thought

Vo=11m/s

V=35m/s

t=0.095s

then I used V=Vo + at

but I was wrong.

hope you can help me with those question

**Thank you

Answer 1

Problem 1: One equation with 2 unknowns is unsolvable. You need some way to find the initial velocity. You have to assume the average is actually the mean. What does negative velocity mean in this case?

Problem 2: You have taken the plug-in approach but have not thought about the values you are using. You have to determine which velocity is negative considering this is a reverse direction. Also, the initial velocity must have the opposite sign of the acceleration. Vo is the initial velocity.

Answer 2

As far as #2 goes, you must take into account the fact that the velocity is changing directions.  Therefore, your V0=35m/s and Vf= -11m/s.  Using Vf=V0 + at and solving for a, you get 35m/s - (-11m/s) / 0.095 s which comes out to 484 m/s.

Answer 3

1) The average velocity is (v + v0)/2 = 35km/hr, v = 0, so v0 = 2 * 35km/hr = 70km/hr

a = (v - v0)/t = (-70km/hr)/7s = -10km/hr/s

   = (-10km/hr/s)(1000m/km)(1hr/3600s) = -2.78 m/s/s

2) Sorry, one question at a time.

Answer 4

1) After landing, a jetliner on a straight runway taxis to a stop at average velocity of -35.0km/hr. If the plane takes 7.00s to come to rest, what are the plane's initial velocity and acceleration?

Ans: -70.0km/hr or -19.4m/s, +2.78m/s^2

Vave = -35 km/h

t =7 sec

Vave = (Vo + Vf)/2

-35= (Vo + 0)/2

-35(2) = Vo

Vo = -70 kph =

Vf = Vo + at

a = (Vf-Vo)/t

a = 0 -(-70)/7

a = (10 km/h)(h/3600 sec)/sec (1000 m/km)

a = 2.78 m/s^2

2)

Vo=11m/s

V=35m/s

t=0.095s

V = Vo + at

11 = -35 + a(0.095)

11 + 35 = a(0.095)

46 = a(0.095)

a = 484.2 m/s^2

a = 4.84 x10^2 m/s^2

Answer 5

1)  Vav = (Vo + Vf)/2 → Vo = 2*(Vave - Vf).  For Vf = 0,

Vo = 2*Vave = 2*35 = 70 km/h = 19.44 m/s.

I would have used the direction the plane is pointing as the + direction, thereby getting +19.4 m/s and -2.78 m/s² for a.

2)  a = dV/dt = (-11 - 35)/.095 = -46/.095 = -482.2 m/s²

Accelerations of cars are comparable to hockey pucks when they (cars) bounce off walls in less than a tenth of a second.....!