Let us look at an example.
A vehicle travels along a curve drug, the speed of the vehicle is uniformly increased from a value
of 20 meters per second to 30 meters per second during a time period of two seconds, the moment the
vehicle travels at a speed of 25 meters per second.
The radius of curvature of the road is two hundred and fifty meters now determine the magnitude of the
vehicle's acceleration at this instant.
So you can see there are two components.
We've got a tangential component.
We've got the normal component, the tangential component.
We will calculate from the increase in velocity over the time that's divided.
And the normal component we know that is calculated by V squared of the road, where is our radius of
curvature?
So we are given all the information that we need.
Let us look.
Are we going to find the answer?
So you're at the top.
I've just drawn a curved road along which the vehicle travels.
And at the instant when the vehicle's velocity is twenty five meters per second, there is a radius
of curvature of two hundred and fifty meters.
So I have drawn that in there and the red line.
So we know the acceleration vector consists of the tangential component and the normal component.
We've got 80 in the tangential direction plus a N in the normal direction a.
That is just DVD t in the tangential direction and the normal component of the acceleration.
That's a V squared over row in the normal direction.
So divided.
Well we are told that the vehicle that the vehicle's velocity increases 120 meters per second to 30
meters per second in a time period of two seconds.
So the change in velocity, that will be 10 seconds.
And that happened over a time period of two seconds.
So that's tangential component of the velocity for that.
Two seconds will be 10 over to.
Change of 10 meters per second in two seconds plus, then our normal component, that is a V squared.
Now it says at the instant where the vehicle travels at twenty five meters per second.
So we have twenty five squared divided by the radius of curvature, which is two hundred and fifty.
This gives us five meters per second squared in the tangential direction, plus two point five meters
per second squared in the normal direction.
I'll note that this is a vector.
We've got direction.
Direction is involved in the question states that we need to find the.
The magnitude of the acceleration, so we need to take the acceleration vector and we need to find the
magnitude of the acceleration.
So how do we do that?
Well, the magnitude of acceleration is the square root of the squeeze of the two components.
We've got the square root of the tangential component plus plus the normal component squared.
That is the square root of five squared, plus 2.5 squared.
That is the square root of thirty one point five.
And that gives us an answer of five point five nine meters per second squared.
Now, this is now a scalar.
It is not any more a vector.
It doesn't indicate direction.
It just tells us what is the magnitude of the acceleration.
So if we add the two components of the acceleration together, what is the total magnitude?
What is the size of the acceleration?
And that is how we would answer an approach.
Such a question.