A car slows down from 23 m/s to rest in a distance of 85 m. What was its acceleration, assumed constant?
To find the acceleration of the car, we can use the following equation:
\(v^2 = u^2 + 2as\)
Where:
v = final velocity (0 m/s, since the car comes to rest)
u = initial velocity (23 m/s)
a = acceleration
s = distance covered (85 m)
Plugging in the given values:
\(0^2 = 23^2 + 2a \cdot 85\)
Simplifying further:
\(0 = 529 + 170a\)
To solve for 'a', we can rearrange the equation:
\(170a = -529\)
Dividing both sides of the equation by 170:
\(a = \frac{-529}{170}\)
Therefore, the acceleration of the car is approximately -3.11 m/s².
To find the acceleration of the car, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (0 m/s, since the car comes to rest),
u is the initial velocity (23 m/s),
a is the acceleration, and
s is the displacement (85 m).
Now, let's rearrange the equation to solve for acceleration:
0^2 = 23^2 + 2a(85)
Simplifying this equation gives us:
0 = 529 + 170a
Subtracting 529 from both sides:
-529 = 170a
Finally, dividing both sides of the equation by 170, we find:
a ≈ -3.11 m/s^2
Thus, the acceleration of the car is approximately -3.11 m/s^2. The negative sign indicates that the car is decelerating or slowing down.
v^2 = 2as
23^2 = 170a
a = -3.11 m/s^2
or,
23t + 1/2 at^2 = 85
23 + at = 0
a = -3.11