Respuesta :
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.
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.
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.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
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:
[tex]v^2 = u^2 + 2ad[/tex]
[tex]0^2 = (475/18)^2 + 2a(0.80)[/tex]
[tex]a = -(475/18)^2 \div 1.6[/tex]
[tex]a \approx -440 ~ m/s^2[/tex]
Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
Answer details
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Tree
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