Find the Total Distance Traveled by an Accelerating Object

[Aromire]

Homework Statement

An object starts from rest with an acceleration of 2.0 m/s2 for 3.0 seconds. It then reduces its acceleration to 1.0 m/s2 for 5.0 additional seconds. The total distance covered is

Homework Equations

The Attempt at a Solution

I've tried using the formula of velocity = acceleration * time but after getting the velocity I was stuck and couldn't go further in getting the solution.

 

[Dick]

Homework Statement

An object starts from rest with an acceleration of 2.0 m/s2 for 3.0 seconds. It then reduces its acceleration to 1.0 m/s2 for 5.0 additional seconds. The total distance covered is

Homework Equations

The Attempt at a Solution

I've tried using the formula of velocity = acceleration * time but after getting the velocity I was stuck and couldn't go further in getting the solution.

You need a formula relating distance to a constant acceleration, intial velocity and time. Don't you have one?

 

[Suraj M]

Do you happen to know any other kinematic equations.
If not try plotting a graph.

 

[Dopplershift]

I might be wrong, but would it be best to solve the two distances by plugging them into the kinematic equations with both accelerations and then adding them together? Of course, for the second distance equation, you would need to find the final velocity of the first part with the acceleration 2.0 m/s^2 and use that as your velocity initial for the second distance equations.

So I would do it like this.

Part 1:
X_f = X₀+V₀t+1/2at²
Where X₀+ V₀t = 0 (assuming we are starting at rest and letting the initial x position be zero) so you can get rid of that part of the equation, so the distance traveled becomes 1/2at²
So X = 1/2(2)*3² = (1/2)* 18 = 9 meters.

Before we start part 2 and add the equations we need to find the V_f for part 1 so we can use that as our V₀ for part 2 since the object doesn't stop and start driving again before it lowers it's acceleration.

We know that V_f = V₀ + at
Well, V₀ = 0 since we started from rest so V_f = a*t
so V_f = 2(3) = 6 m/s

For part 2: we can use the same distance equation:
X_f = X₀+V₀t+1/2at²

Where X₀ is = 9 meters and V₀ = 6m/s

So we have 9 meters + 6m/s*t+ 1/2at²

Well, we know that t is equal to 5 seconds and acceleration is 1 m/s, so we have

9 meters + 6m/s*(5)+ (1/2)*(1)+(5²) which is equal to 51.5 meters which should be your answer

 

[Suraj M]

which is equal to 51.5 meters which should be your answer

We are very happy to have on PF,
But we have a few rules and regulations for helpers, i know the feeling you might get when you see a question which you can solve, the urge to solve will be there, i had it too. But it's always more useful to the person trying to solve the problem to do it on his own, Giving a few hints on how he should solve the questin will help him more than giving him the whole answer. Thank you for participating in PF, and keep doing so., after reading this:https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/
Thank you

 

[Aromire]

Thank you very much Dopplershift, you are a life saver and I actually spent hours trying to figure it out.

I might be wrong, but would it be best to solve the two distances by plugging them into the kinematic equations with both accelerations and then adding them together? Of course, for the second distance equation, you would need to find the final velocity of the first part with the acceleration 2.0 m/s^2 and use that as your velocity initial for the second distance equations.

So I would do it like this.

Part 1:
X_f = X₀+V₀t+1/2at²
Where X₀+ V₀t = 0 (assuming we are starting at rest and letting the initial x position be zero) so you can get rid of that part of the equation, so the distance traveled becomes 1/2at²
So X = 1/2(2)*3² = (1/2)* 18 = 9 meters.

Before we start part 2 and add the equations we need to find the V_f for part 1 so we can use that as our V₀ for part 2 since the object doesn't stop and start driving again before it lowers it's acceleration.

We know that V_f = V₀ + at
Well, V₀ = 0 since we started from rest so V_f = a*t
so V_f = 2(3) = 6 m/s

For part 2: we can use the same distance equation:
X_f = X₀+V₀t+1/2at²

Where X₀ is = 9 meters and V₀ = 6m/s

So we have 9 meters + 6m/s*t+ 1/2at²

Well, we know that t is equal to 5 seconds and acceleration is 1 m/s, so we have

9 meters + 6m/s*(5)+ (1/2)*(1)+(5²) which is equal to 51.5 meters which should be your answer

I might be wrong, but would it be best to solve the two distances by plugging them into the kinematic equations with both accelerations and then adding them together? Of course, for the second distance equation, you would need to find the final velocity of the first part with the acceleration 2.0 m/s^2 and use that as your velocity initial for the second distance equations.

So I would do it like this.

Part 1:
X_f = X₀+V₀t+1/2at²
Where X₀+ V₀t = 0 (assuming we are starting at rest and letting the initial x position be zero) so you can get rid of that part of the equation, so the distance traveled becomes 1/2at²
So X = 1/2(2)*3² = (1/2)* 18 = 9 meters.

Before we start part 2 and add the equations we need to find the V_f for part 1 so we can use that as our V₀ for part 2 since the object doesn't stop and start driving again before it lowers it's acceleration.

We know that V_f = V₀ + at
Well, V₀ = 0 since we started from rest so V_f = a*t
so V_f = 2(3) = 6 m/s

For part 2: we can use the same distance equation:
X_f = X₀+V₀t+1/2at²

Where X₀ is = 9 meters and V₀ = 6m/s

So we have 9 meters + 6m/s*t+ 1/2at²

Well, we know that t is equal to 5 seconds and acceleration is 1 m/s, so we have

9 meters + 6m/s*(5)+ (1/2)*(1)+(5²) which is equal to 51.5 meters which should be your answer