How fast will I be travelling after 10 seconds if I start from standstill and accelerate at a rate of 1m.s-2?
Nov 16
]sara[ asked:
Calculate the velocity of an object at any time given its acceleration.
How fast will I be travelling after 10 seconds if I start from standstill and accelerate at a rate of 1m.s-2?
Calculate the velocity of an object at any time given its acceleration.
How fast will I be travelling after 10 seconds if I start from standstill and accelerate at a rate of 1m.s-2?
Please explain in detail so I can understand easily.
The most explanatory and correct answer will get 10 points.
Thanks.
Tour guide
well i did a lil calculus in high school so here goes acceleration is the derivative of velocity so find the antiderivative and u find rate of velocity which would equal 1x since the acceleration equals just 1 constant there should be more info because once u have the antiderivative 1x there is another constant to be added on there x+ C idk thats just how we did it in calculus class
Do you understand what ‘accelerate’ means? That’s the simplest problem you’ll ever encounter. You have only two numbers to work with here.
Something happens once per second for ten seconds, how many times does it happen? Three? Come on.
What happens is it speeds up by one meter per second. So what’s ten times one meter per second (or anything else)? Ten meters per second, do you suppose?
I think you understand more than you realize you do, if only you don’t make a simple problem complicated.
you are talking about a constant acceleration.
First, let’s define initial velocity v0, and acceleration a(t)
Now, v(t) = Int_(0,t) a(s) ds + v0
So if a(t) = 1 m/s^2, then this integral just goes to
v(t) = Int_(0, 10) 1 ds + v0 = 10 + v0
But the thing is starting from standstill, so v0 = 0, so thus the final velocity at t=10 seconds will be 10 m/s
Now, the other way to do this is:
v = v0 + a*t
this comes from the same equation, but is just a simpler algebraic statement of it given that acceleration is a constant (when acceleration is a non-constant function of t, you need to actually integrate the thing to find out its overall effect on velocity, but in the case of a constant acceleration, things simplify quite a bit)
In 1-dimensional kinematics, there are a few equations to remember:
Define v to be final velocity, v0 initial velocity, x to be final position, x0 initial position, and a acceleration…v, x, and a are functions of t, while x0 and v0 are fixed according to the problem statement.
Equations to remember:
x = x0 + v0*t + a*t^2 / 2
v^2 – v0^2 = 2a*(x-x0)
v = v0 + a*t
The first and last come from integrating the equation
d^2/dt^2 x(t) = a
The third comes from integrating the equation and then making a substitution from the third equation into the first and simplifying things out a bit.
Acceleration means any change in velocity, which means a change in speed or direction or both. In your simple case of straight-line motion, the only thing changing is the speed, which increases by one meter per second for each second of time. So, after one second your speed will be one meter/second, after two seconds it will be two meters/second, and after ten seconds it will be ten meters/second – a constant acceleration produces a changing velocity (calculus is not required to understand the concept but it is useful in solving real-world problems).
You square the time, 10 X 10 = 100, X 1 = 100 meters per second.