Question

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?

Answer 1

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.

Answer 2

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

Further explanation

Acceleration is rate of change of velocity.

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

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

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

This problem is about Kinematics.

We will solve it in the following way.

Given:

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

Unknown:

acceleration = a = ?

Solution:

$v^2 = u^2 + 2ad$

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

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

$a \approx -440 ~ m/s^2$

Learn more

• 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

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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