# Kinematics in 1 dimension

• future_vet
In summary, The ball reaches a height of 11 meters and has a velocity of 0 m/s at the apex after being thrown vertically upward with an acceleration of -9.8 m/s^2.

#### future_vet

Hello! Could you please check for errors in the following HW problems?
Thank you!

A.

## Homework Statement

A rolling ball moves from x1=3.4 cm to x2=-4.2 cm during the time from t1=3.0 s to t2=6.1 s. What is the average velocity?

and

## The Attempt at a Solution

Average velocity = (x2 - x1)/(t2-t1) ~ (-2.5 cm/s)

B.

## Homework Statement

A sports car accelerates from rest to 95 km/h in 6.2 s. What is the average acceleration in m/s^2?

and

## The Attempt at a Solution

average acceleration = (v2 - v1)/(t2-t1) ~4.3 m/s^2.

C.

## Homework Statement

At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s^2. At this rate, how long does it take to accelerate from 80km/h to 110 km/h?

and

## The Attempt at a Solution

v1= 80 km/h = 22 m/s
v2= 110 km/h = 31 m/s
Average acceleration = 1.6m/s^2=(v2 - v1)/(∆t)=> it takes 5.6 seconds.

D.

## Homework Statement

A horse canters away from its traiener in a straight line, moving 116m away in 14.0 seconds. It then turns abruptly and gallops halfway back in 4.8 seconds. Calculate its average speed and its average velocity for the entire trip, using away from the trainer as the positive direction.

and

## The Attempt at a Solution

Average speed=d/t=(116 + 58)/(14.0 + 4.8 seconds) = 9.3 m/s.
Average velocity= ∆x/∆t=3.1m/s.

E.

## Homework Statement

A light plane nust reach a speed of 33m/s for takeoff. How long a runway is needed if the constant acceleration is 3.0 m/s^2?

and

## The Attempt at a Solution

(x-xo) = (v^2 -vo^2)/2a = (33m/s)^2/6.0= 181 meters.

F.

## Homework Statement

In coming to a stop, a car leaves skid marks 92 m long on the highway. Assuming a decelration of 7.00m/s^2, estimate the speed of the car just before braking.

and

## The Attempt at a Solution

V= √(2x7.00x92) ~ 36 m/s.

G.

## Homework Statement

Estimate how long it took King Kong to fall straight down from the tip of the Empire State building (380 m high) and his velocity just before landing.

and

## The Attempt at a Solution

y=vot1 + 1/2 x gt
t= 8.8 seconds.

V^2=Vo^2 + 2gh
V = 86 meters/seconds.

H.

## Homework Statement

A ballplayer catches a ball 3.0 seconds after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

and

## The Attempt at a Solution

I don't know which equation to use here... Could you please tell me which one it is?

Thanks!

Correct equation for B (if that's how your teacher gives you the equations)... How'd you get that answer? (A seems correct to me.)
Ditto for C. You have everything you need, including an equation that would work, but I can't figure out where you're getting that answer.

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DEFG are all fine.

Hint for H: when he throws the ball, it's going up for half the time, and coming down for half the time. Split the motion in half and look at either half.

future_vet said:
Hello! Could you please check for errors in the following HW problems?
Thank you!

A.

## Homework Statement

A rolling ball moves from x1=3.4 cm to x2=-4.2 cm during the time from t1=3.0 s to t2=6.1 s. What is the average velocity?

and

## The Attempt at a Solution

Average velocity = (x2 - x1)/(t2-t1) ~ (-2.5 cm/s)
Correct

B.

## Homework Statement

A sports car accelerates from rest to 95 km/h in 6.2 s. What is the average acceleration in m/s^2?

and

## The Attempt at a Solution

average acceleration = (v2 - v1)/(t2-t1) ~4.3 m/s^2.
edit: yup, this is right; I didn't notice your change of units either!
C.

## Homework Statement

At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s^2. At this rate, how long does it take to accelerate from 80km/h to 110 km/h?

and

## The Attempt at a Solution

v1= 80 km/h = 22 m/s
v2= 110 km/h = 31 m/s
Average acceleration = 1.6m/s^2=(v2 - v1)/(∆t)=> it takes 5.6 seconds.
Correct

edit: late!

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edit: oh, never mind... carelessness on my part.
I wasn't paying careful enough attention to your units; shame on me!
(km/hr vs. m/s)
You were correct.

(You caught both of us off guard :) Both of the answers I said I didn't see where you got the solution from were in fact correct.) I rushed through your calculations a little too quickly.

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Hi!

So everything is correct up to G then? Thanks for your help :)

Let's see for H:
The ball is going to reach the apex after 1.5 seconds, where the velocity will be 0.
a=9.8 m/s^2, but will be negative on the way down...
We don't have the velocity
We don't have the height
We don't have the speed... but we know that the speed will be zero at the apex.

I have to say I'm still lost... It doesn't seem complicated at first, but I am not sure what to do here...

=/

future_vet said:
Hi!

So everything is correct up to G then? Thanks for your help :)

Let's see for H:
The ball is going to reach the apex after 1.5 seconds, where the velocity will be 0.
a=9.8 m/s^2, but will be negative on the way down...
We don't have the velocity
We don't have the height

You can calculate the height (at least the height relative to the point at which he throws the ball upwards, which I assume is what is required) but considering the point at which v=0 (hint: kinematic equations; more specifically, the one involving d, v, t and a).

Now you have h, you can calculate v.
We don't have the speed... but we know that the speed will be zero at the apex.
Speed is magnitude of velocity. Since the ball is thrown vertically, the velocity is equal to the speed here.

d = v1 x t + 1/2 x a x t^2
d= 0 x 1.5 seconds + 1/2 x 9.8 x (1.5)^2
d= 11 meters.

Would this be correct for the height?

Thanks..

future_vet said:
d = v1 x t + 1/2 x a x t^2
d= 0 x 1.5 seconds + 1/2 x 9.8 x (1.5)^2
d= 11 meters.

Would this be correct for the height?

Thanks..

That's correct, although your original equation should read d=vt-1/2at2. a=-9.8, and so this becomes your second line.

For the height:
d=0x1.5 + 1/2 x (-9.8) x (1.5)^2 then?

From there, I should have all I need to find the speed?

I had no problem doing the homework up to this point, but for some reason it's very confusing to me =/

Thanks!

future_vet said:
For the height:
d=0x1.5 + 1/2 x (-9.8) x (1.5)^2 then?

From there, I should have all I need to find the speed?

I had no problem doing the homework up to this point, but for some reason it's very confusing to me =/

Thanks!

you have calculated the height above (using d=vt-1/2at2). Now, you can use a similar equation for initial velocity, and consider the whole journey of the ball: d=ut+1/2at2. You know the time for the whole flight, a, and d is the value you have just calculated, so you can solve for u (the initial velocity)

Wit a= -9.8, I get a negative height, and then a negative speed...

Here's what I did:
d=vt-1/2at^2
d= 0 x 1.5 + 0.5 x 9.8 x 1.5^2 = - 11 meters

For the speed:
d (above) = vt-1/2at^2 = v x 3 seconds - 0.5 x (-9.8) x 3^2

What am I doing wrong?...

Thank you and sorry for all the trouble :)

future_vet said:
Wit a= -9.8, I get a negative height, and then a negative speed...

Here's what I did:
d=vt-1/2at^2
d= 0 x 1.5 + 0.5 x 9.8 x 1.5^2 = - 11 meters

This isn't correct. This d= 0 x 1.5 + 0.5 x 9.8 x 1.5^2 is correct, but it does not equal -11; it is equal to 11!

Note that the original equation has the term -1/2at^2 in it, but when you substitute in a=-9.8, this term becomes + 0.5 x 9.8 x 1.5^2 (like you did above). The correct answer is 11m.

For the initial speed, you must use the equation d=ut+1/2at^2 (note the difference in sign between the equation for initial, and final, velocity). Sub in what you know and you should be fine.

So, I think I got it now..

For the height, you were right, I apologize for that!

For the speed:
11 = 3v + 1/2 x -9.8 x 3^2 (I use the whole 3 seconds for the trajectory, not half the seconds, right?)
3v = 55.1
v= 18 meters per second.

Good?

future_vet said:
So, I think I got it now..

For the height, you were right, I apologize for that!

That's ok; it's more important that you recognise the difference between the two equations, than you make a simple algebraic mistake.

For the speed:
11 = 3v + 1/2 x -9.8 x 3^2 (I use the whole 3 seconds for the trajectory, not half the seconds, right?)
3v = 55.1
v= 18 meters per second.

Good?

Yup, that's correct.

Thank you!

## 1. What is kinematics in 1 dimension?

Kinematics in 1 dimension is the branch of physics that studies the motion of objects along a straight line. It involves analyzing the position, velocity, and acceleration of an object as it moves in a single direction.

## 2. What are the basic equations used in kinematics in 1 dimension?

The basic equations used in kinematics in 1 dimension are the displacement equation (Δx = xf - xi), the average velocity equation (v = Δx/Δt), the average acceleration equation (a = Δv/Δt), and the final velocity equation (vf = vi + at).

## 3. How does acceleration affect an object's motion in 1 dimension?

Acceleration affects an object's motion in 1 dimension by changing its velocity. If an object has a positive acceleration, it will increase in speed, while a negative acceleration will cause it to decrease in speed. If the acceleration is zero, the object will maintain a constant velocity.

## 4. Can an object's position change while its velocity remains constant?

Yes, an object's position can change while its velocity remains constant. This is known as uniform motion and occurs when the object is moving at a constant speed in a straight line. However, the direction of the object's motion may change, resulting in a change in position.

## 5. How can kinematics in 1 dimension be applied to real-world situations?

Kinematics in 1 dimension can be applied to real-world situations such as calculating the distance traveled by a car, determining the speed of a falling object, or predicting the position of a projectile. It is also used in fields like engineering, sports, and transportation to analyze and improve the motion of objects.