Question

HELP PLEASE

Your teacher's car can go from rest to 25 m/s (≈55 mph) in 10 seconds. The car's velocity changes at a uniform rate. Below is a quantitative velocity vs time graph that represents the motion of the car. Determine the acceleration of the car.

Answer 1

Recall this kinematic equation:

a =
`(Vi+Vf)/(Δt)`

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 =
`(0+25)/(10)`

a = 2.5 m/s²

Answer 2

2.5m/s^2

Step-by-step 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.