Question

A car slows down from 23 m/s to rest in a distance of 85 m. what was its acceleration, assmed constant

Answer 1

First calculate the time by using following equation:
d= (vi+vf)/2 x t 

where t = time

vi is the initial velocity = 23 m/s

vf is the final velocity  = 0 m/s

and d is the distance covered = 85m

85 = (23 + 0)/2 x t

t = 85/11.5 = 7.39s

Now use this equation to find acceleration:

a= (vf-vi)/t 

a = 0 – 23/ 7.39

a = -3.112 m/s²

so, the acceleration is 3.112 m/s²

Answer 2

The magnitude of the car's acceleration was 3.1 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!

Given:

initial velocity = u = 23 m/s

final velocity = v = 0 m/s

distance = d = 85 m

Unknown:

acceleration = a = ?

Solution:

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

$0^2 = 23^2 + 2a(85)$

$0 = 529 + 170a$

$170a = -529$

$a = -529 / 170$

$\large {\boxed {a = -3.1 ~ m/s^2} }$

The acceleration was 3.1 m/s² in the opposite direction to its velocity.

A negative sign means the direction of acceleration is opposite to the direction of velocity.

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