Calvin and hobbes go for a sleigh ride.

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SUMMARY

The discussion focuses on calculating the time taken for Calvin and Hobbes to complete a sleigh ride from Point A to Point D along a frictionless path. The ride consists of three segments: a downhill slope (AB) of 30m height and 40m length, a level surface (BC) of 60m, and an uphill slope (CD) of 30m height and 30m length. The time taken from A to B is calculated as 4.1 seconds, and from B to C as 2.5 seconds. The challenge remains in determining the time for segment CD, which requires applying energy conservation principles and solving for acceleration and time.

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  • Understanding of kinematics, specifically the equations of motion.
  • Knowledge of energy conservation principles in physics.
  • Ability to analyze free body diagrams and calculate forces.
  • Familiarity with quadratic equations for solving motion problems.
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  • Study kinematic equations for motion under constant acceleration.
  • Explore free body diagram analysis for inclined planes.
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This discussion is beneficial for physics students, educators, and anyone interested in solving mechanics problems involving motion on inclined planes and energy conservation.

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Homework Statement



Calvina nd Hobbes go for a sleigh ride. They start from rest at Point a and proceed to point D along the path ABCD. AB is a downhill slope of height 30m and length 40 m (thus hypotenuse where they ride on is 50m), BC is level surface 60m and CD is an uphill slope of hieght 30m and length 30m. The surfaces are all frictionless. Calculate how long the trip from A to D took.

Homework Equations



Δd= v1 Δt +0.5a *Δ t^2


The Attempt at a Solution


acceleration from A to B is equal to gsintheta from free body diagram. theta is 37 degrees.
using equation above, time for A to B is 4.1 seconds.

B to C is constant velocity because net force is 0. therefore, time is 60 divided by final velocity of AB which is 24m/s. therefore 60/24 is 2.5

Then I am stuck on time for CD.
If anyone can help, thanks so much :)

 
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Hello, and welcome to PF!

Can you find the acceleration between C and D?
 
What you did was correct.

So use your formula for d again. What is d this time? what is v1? and what is a? This time you need to solve a quadratic since v1 is not zero the way it was at A.

You can avoid the quadratic by realizing v(t) between C and D is v1 + a*t. Let T be the time going from C to D, then v(T) = V(D) = v1 + a*T. Remember a is negative. What would V(D) be? Use energy conservation in going from A to D to answer that question, then solve for T.