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
The magnitude of the car's acceleration was 3.1 m/s²
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!
Given:
initial velocity = u = 23 m/s
final velocity = v = 0 m/s
distance = d = 85 m
Unknown:
acceleration = a = ?
Solution:
[tex]v^2 = u^2 + 2ad[/tex]
[tex]0^2 = 23^2 + 2a(85)[/tex]
[tex]0 = 529 + 170a[/tex]
[tex]170a = -529[/tex]
[tex]a = -529 / 170[/tex]
[tex]\large {\boxed {a = -3.1 ~ m/s^2} }[/tex]
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.
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
