Recall this kinematic equation:
a = [tex]\frac{Vi+Vf}{Δt}[/tex]
This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).
a is the acceleration.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 25 m/s.
Δt = 10 s
Substitute the terms in the equation with the given values and solve for a:
a = [tex]\frac{0+25}{10}[/tex]
Answer:
2.5m/s^2
Explanation:
We have the first equation of motion here:
V(f) = V(i) + at (equation 1)
Now,
V(f) = Final velocity = 25 m/s
V(i) = Initial velocity = 0 m/s (Since car starts moving from rest, so intial velocity has to be considered as 0 m/s)
a = Acceleration = ? (It has to be determined)
t= Time = 10 seconds
Now by putting values in equation 1
25 = 0 + (a)*(10)
25 = 10*a
a = 25/10
a = 2.5 m/s^2
Additionally, remember that we have three equations of motion
First equation is V(f) = V(i) + at
Second is S = V(i)t + (1/2)a(t^2)
Third is 2as = [V(f)]^2 - [V(i)]^2
Where S = Distance
t = Time
a= Acceleration
V(f) = Final velocity
V(i) = Initial velocity
We always decide to use any of these equations according to the data given, like you can see that I used first equation because it is independent of Distance, since in data, there was no distance given, so i used first equation of motion.
