A car accelerates from 5 m/s to 21 m/s at a constant 3.0 m/s 2 . How far does it travel while accelerating?

guys. Let's get some practice. Go ahead and pause the video if you haven't already done so and give this one a shot. So we have a constant 3 meters per second squared that the car this car is accelerating through. We're asked to find how far it travels while accelerating. So we've got this list of steps here. We're just gonna go ahead and follow them. First we gotta do is just draw a quick little diagram. So we got the car that is accelerating horizontally like this. And let's just say the car looks like that. That's pretty good. Right. So we've got an initial velocity that says v naught is equal to 5 meters per second. It accelerates, and then finally, it's traveling at 21 when it gets to the end. So later on here, what happens is that the velocity has increased and it's now 21. We know that the acceleration throughout this motion over here, this a here, is equal to 3. So now now we just go ahead and list the variables. We've got delta x, v naught, v final, a, and t. So we know what the v naught is. Whoops. That's actually 5. We know that the final velocity is 21. We know the acceleration is 3. So that's the diagram listing the variables. We've already identified the known variables. What's the target? What are we actually looking for? Well, how far indicates a distance or displacement, and that's going to be our delta x. So this is really what we're trying to find here. So that is the second step, identifying the target. Now, we just go ahead and pick the UAM equation without the ignored variable. And so the ignored variable is the one that we're not given and we're not asked for. So if you look at this, we've got 3 out of 5 variables, so we can use our equations. We're looking for this delta x, which means that our time is actually going to be the ignored variable, sad face. So now we just go and pick the equation that does not have time. And if you look here, this time is in equation 1, so we can't use that. So let me go ahead and write that. So we've got time, so we can't use equation 1. Time, we can't use equation 3. And time, we can't use equation 4. So that leaves only equation number 2. So that means that we're gonna use equation number 2, which says that v squared is equal to v initial squared plus 2a times delta x. And so here is our target variable. You'll notice that we have everything else to solve the problem. We know that v is 21 squared, v naught is 5, so that's gonna square. And then we're gonna have 2 times the acceleration, which is 3 times delta x. So you could just go ahead and square all of these numbers and just work with the numbers, but I'm just gonna wait to do all the calculations at the end. So I can move the 5 squared over, so I'm gonna get 21 squared minus 5 squared. And now I get equals, this is 2 times 3, so this becomes 6 times delta x. And now I just divide by 6. So I get 21 squared minus 5 squared, divided by 6 is it delta x? And if you go ahead and work this out, just plug it in once, you're gonna get 69.3 meters. And that's our answer. Alright, Alright, guys. Let me know if you have any questions. That's it for this

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2. 1D Motion / Kinematics

Kinematics Equations

Physics

2. 1D Motion / Kinematics

Kinematics Equations

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Multiple Choice

A car accelerates from 5 m/s to 21 m/s at a constant 3.0 m/s². How far does it travel while accelerating?

437 m

2.67 m

69.3 m

42.7 m

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Identify the known values: initial velocity \( v_i = 5 \text{ m/s} \), final velocity \( v_f = 21 \text{ m/s} \), and acceleration \( a = 3.0 \text{ m/s}^2 \).

Use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance: \( v_f^2 = v_i^2 + 2a d \), where \( d \) is the distance traveled.

Rearrange the equation to solve for \( d \): \( d = \frac{v_f^2 - v_i^2}{2a} \).

Substitute the known values into the equation: \( d = \frac{(21)^2 - (5)^2}{2 \times 3.0} \).

Calculate the expression to find the distance \( d \).

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A car accelerates from 5 m/s to 21 m/s at a constant 3.0 m/s2. How far does it travel while accelerating?

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Kinematics Equations practice set

A car accelerates from 5 m/s to 21 m/s at a constant 3.0 m/s². How far does it travel while accelerating?