Answer:
-2.79 m/s²
Explanation:
Given:
v₀ = 20 m/s
v = 11 m/s
Δx = 50 m
Find: a
v² = v₀² + 2aΔx
(11 m/s)² = (20 m/s)² + 2a (50 m)
a = -2.79 m/s²
Round as needed.
Explanation:
Given that,
Initial speed of the vehicle, u = 45 mi/h = 20 m/s
The final speed of the vehicle, v = 25 mi/h = 11 m/s
The speed limit will change in 50 m i.e. the distance is, d = 50 m
To find,
The acceleration of the vehicle that is needed to reach the new speed limit before it begins.
Solution,
Let a is the acceleration of the vehicle. It can be calculated using third equation of motion i.e.
[tex]v^2-u^2=2ad[/tex]
Solving for a.
[tex]a=\dfrac{v^2-u^2}{2d}[/tex]
Put all the values,
[tex]a=\dfrac{11^2-20^2}{2(50)}\\\\a=-2.79\ m/s^2[/tex]
Answer:
The acceleration of the vehicle is [tex]2.79\ m/s^2[/tex] and it is deaccelerating.
Reference:
https://brainly.com/question/12550364
