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irinina [24]
3 years ago
7

A car traveling 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m

. Part A What was the magnitude of the average acceleration of the driver during the collision?

Physics
2 answers:
alexdok [17]3 years ago
8 0
Assuming the acceleration is constant, then we can use the derived equations for rectilinear motion. The equation is written below:

2ax = v²- v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity

Since the car came to a stop, v = 0. Substituting the values,
2a(0.80 m) =0² -  [(95 km/h)(1000 m/1 km)(1 h/3600 s)]²
Solving for a,
a = -435.23 m/s²
The sign is negative because it is decelerating.
IceJOKER [234]3 years ago
7 0

The magnitude of the average acceleration of the driver during the collision is about 440 m/s²

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

This problem is about Kinematics.

We will solve it in the following way.

<u>Given:</u>

initial velocity of the car = u = 95 km/h = (475/18) m/s

distance = d = 0.80 m

final velocity of the car = v = 0 m/s

<u>Unknown:</u>

acceleration = a = ?

<u>Solution:</u>

v^2 = u^2 + 2ad

0^2 = (475/18)^2 + 2a(0.80)

a = -(475/18)^2 \div 1.6

a \approx -440 ~ m/s^2

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Tree

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       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

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        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

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       D = \frac{15*v_{0} ^{2} }{2*g}

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