A car is 12 m from the bottom of a ramp that is 8.0m long at its base and 6.0m high at its tallest.The car moves from rest toward the ramp with an acceleration of magnitude 2.5 m/s?. At someinstant after the car starts moving, a crate is released from rest from some position along theramp. The crate and car reach the bottom of the ramp at the same moment and traveling at thesame speed. The crate accelerates along the ramp at a rate of 5.88 m/s².2.5 m/s?6.0 m8.0 m12 m2015 Pearson Education InGDraw a pictorial representation first (Tactics Box 1.5). Refer to the Problem Solving Strategies(General Pg21 and 2.1, 4.1) and Models 2.1, 2.2, and 4.1 for best Problem Solving Strategies. Hint:Use two different coordinate systems for the car (x) and the crate (s).A. What is the speed of the car when it reaches the base of the ramp?B. What is the distance, d up the ramp, at which the crate was released?C. How many seconds after the car started was the crate released?

Name: A car is 12 m from the bottom of a ramp that is 8.0m long at its base
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Description: A car is 12 m from the bottom of a ramp that is 8.0m long at its base and 6.0m high at its tallest.The car moves from rest toward the ramp with an acceleration of magnitude 2.5 m/s?. At someinstant after the car starts moving, a crate is released fr

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A car is 12 m from the bottom of a ramp that is 8.0m long at its base and 6.0m high at its tallest.
The car moves from rest toward the ramp with an acceleration of magnitude 2.5 m/s?. At some
instant after the car starts moving, a crate is released from rest from some position along the
ramp. The crate and car reach the bottom of the ramp at the same moment and traveling at the
same speed. The crate accelerates along the ramp at a rate of 5.88 m/s².
2.5 m/s?
6.0 m
8.0 m
12 m
2015 Pearson Education InG
Draw a pictorial representation first (Tactics Box 1.5). Refer to the Problem Solving Strategies
(General Pg21 and 2.1, 4.1) and Models 2.1, 2.2, and 4.1 for best Problem Solving Strategies. Hint:
Use two different coordinate systems for the car (x) and the crate (s).
A. What is the speed of the car when it reaches the base of the ramp?
B. What is the distance, d up the ramp, at which the crate was released?
C. How many seconds after the car started was the crate released?

Expert Solution

Step 1

Given:

a. Speed of car (at the base of the ramp)

Here the car starts from rest, thus its initial velocity, u = 0 m/s

Now, when it reaches the bottom of the ramp where distance, s = 12 m

and acceleration, a = 2.5 m/s²

Consider the kinematic equation:

$\begin{array}{rcl} v^{2} & = & u^{2} + 2 as \\ Substitutitng values, we get: \\ v^{2} & = & {(0 m / s)}^{2} + 2 (2.5 m / s^{2}) (12 m) \\ = & 60 m^{2} / s^{2} \\ v & = & \sqrt{60 m^{2} / s^{2}} \\ = & 7.7454 m / s \\ \approx & 7.70 m / s \end{array}$

Now, the time taken by the car to reach the bottom of the ramp will be :

$\begin{array}{rcl} v & = & u + at \end{array}$

Substituting the values in above equation, we get:

$\begin{array}{rcl} (7.70 m / s) & = & 0 + (2.5 \times t) \end{array}$

$\begin{array}{rcl} Re-arranging the equation and solving for t, we get: \\ t & = & (\frac{7.70 m / s}{2.5 m / s^{2}}) \\ = & 3.09 s \end{array}$

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