Atruck with an initial speed of 12 meters per second accelerates uniformly at 2.0 meters per second2 for 3.0 seconds.  what is the total distance traveled by the truck during this 3.0-second interval?
1. 9.0
2. 25
3. 36
4. 45

The total distance traveled by the truck is 45 meters ⇒ answer 4

Explanation:

A truck with an initial speed of 12 meters per second accelerates

uniformly at 2.0 meters per second² for 3.0 seconds

The initial speed is 12 m/s

The acceleration is 2 m/s²

The time is 3 seconds

We need to find the total distance traveled by the truck during this time

interval

We need rule of distance has initial speed , acceleration and time

So we will use the rule , where u is the

initial speed, t is the time, and a is the acceleration

u = 12 m/s , a = 2 m/s² , t = 3 s

s = 36 + 9 = 45 meters

The total distance traveled by the truck is 45 meters

We want to find the total distance traveled by a truck during a constant acceleration.

We will see that the correct option is the last one, the distance traveled is 45 meters.

We know that acceleration is defined as the rate of change of the velocity.

So if the initial speed of the truck is 12 m/s and it accelerates at 2 m/s^2, the velocity equation of the truck will be:

v(t) = (2 m/s^2)*t + 12m/s    for 0s < t < 3s

To get the position equation we need to integrate over time, we will get:

p(t) = (1/2)*(2 m/s^2)*t^2 + (12m/s)*t + p₀

Where p₀ is the initial position of the truck.

Now we want to find how much distance is traveled in the 3 seconds interval when the truck accelerates, then we just need to compute:

p(3s) - p(0s) =  (1/2)*(2 m/s^2)*(3s)^2 + (12m/s)*3s = 45m

Then the correct option is the last one, the distance traveled is 45 meters.

If you want to learn more, you can read:

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the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever