Could anyone explain to me how to solve this basic acceleration problem?

Question

A runner achieves a velocity of 11.1m/s 9s after he begins. I found his acceleration to be 1.23m/s/s. The next question asked,"What distance did he cover?" I'm not sure I understand the concept of meters per second per second. A basic explanation would be helpful, considering I only started physics two weeks ago. Thanks so much!

Answer 1

Think of it this way.

Meters per second is the rate at which you change distance over time.



Velocity per second is the rate at which you change speed over time.



The units of velocity are meters/second. And if you are changing velocity, the units are

velocity / second or (meters/sec) /sec



If you are driving a car at contstant speed, the speedometer needle doesnt move. It stays fixed at some particular velocity.



If you accelerate, you see the speedometer needle moving. The velocity of the needle is the rate of change of the velocity of the car, or the acceleration of the car.



To figure out the distance the runner travels given the acceleration you need to integrate acceleration to get velocity as a function of time, and then integrate the velocity expression to get distance as a function of time.



let



a = acceleration

v=velocity

x = distance





dv/dt = a, or dv = a dt

integrating,



v =a * t



but,

dx / dt = v



or,

dx / dt = a * t



rearranging,



dx = a * t * dt



Integrate this equation to get



x = 1/2 * a * t^2



Plug values of a and t into the equation to calculate the distance the accelerating runner travels over that period of time.



Hope this helps,



-Guru

Answer 2

It is sometimes easier to think of m/s/s as: m/s every second



In other words, acceleration is a measure of how the velocity (m/s) is changing. If you are going 10 m/s and staying steady at that speed, then your acceleration is 0 m/s each second.



If you are speeding up by 1 m/s each second, then you are accelerating at 1 m/s/s and if you started at a steady 10 m/s and started accelerating at 1 m/s each second, then after a second you are going 11 m/s and after 2 seconds 12 m/s.



Just like the problem here. You correctly calculated that if you start at zero speed (stopped) and you accellerate at 1.23 m/s for 9 seconds you will be going 11.1 m/s at that time.



Calculating the distance traveled over that time is a little tricky because the runner is going a different speed every microsecond of the time. You have to add up the distance covered each instant over the time all at different speeds. It requires mathematics called calculus to make that kind of summation. Fortunately for people who haven't learned a lot of calculus, the calculus has been done ahead of time of all the common situations and equations have been created.



The one you need for this is:



d=at^2



This equation says that the distance traveled by something under constant acceleration is equal to that acceleration times the square of the time of the acceleration.



For your problem that means:



d=(1.23)m/s/s(9 sec)^2

d=1.23*81 m/sec^2 sec^2

d=99.9 m



This is a very fast runner. He ran the 100 meter dash (nearly) in 9 seconds.



keep at it

math is power

Answer 3

V= 11.1m/s

T= 9s

D= ?



D = (½)(A)(T)²

D = (½)(1.23m/s²)(9s)²



now just punch it into a calculator.



(0.5 * 1.23 * 9) ²



its not too difficult, to find the distance of an object moving at constant acceleration the equation will be the one i showed above. if it helps you to understand acceleration you could always write it as X m/s².

Answer 4

if people were actually as smart as they claim then they would explain things in words you can understand rather than speaking a different language. unfortunately the world is stupid = \

Answer 5

no