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Let us look at an example.

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A vehicle travels along a curve drug, the speed of the vehicle is uniformly increased from a value

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of 20 meters per second to 30 meters per second during a time period of two seconds, the moment the

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vehicle travels at a speed of 25 meters per second.

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The radius of curvature of the road is two hundred and fifty meters now determine the magnitude of the

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vehicle's acceleration at this instant.

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So you can see there are two components.

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We've got a tangential component.

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We've got the normal component, the tangential component.

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We will calculate from the increase in velocity over the time that's divided.

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And the normal component we know that is calculated by V squared of the road, where is our radius of

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curvature?

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So we are given all the information that we need.

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Let us look.

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Are we going to find the answer?

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So you're at the top.

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I've just drawn a curved road along which the vehicle travels.

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And at the instant when the vehicle's velocity is twenty five meters per second, there is a radius

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of curvature of two hundred and fifty meters.

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So I have drawn that in there and the red line.

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So we know the acceleration vector consists of the tangential component and the normal component.

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We've got 80 in the tangential direction plus a N in the normal direction a.

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That is just DVD t in the tangential direction and the normal component of the acceleration.

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That's a V squared over row in the normal direction.

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So divided.

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Well we are told that the vehicle that the vehicle's velocity increases 120 meters per second to 30

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meters per second in a time period of two seconds.

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So the change in velocity, that will be 10 seconds.

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And that happened over a time period of two seconds.

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So that's tangential component of the velocity for that.

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Two seconds will be 10 over to.

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Change of 10 meters per second in two seconds plus, then our normal component, that is a V squared.

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Now it says at the instant where the vehicle travels at twenty five meters per second.

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So we have twenty five squared divided by the radius of curvature, which is two hundred and fifty.

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This gives us five meters per second squared in the tangential direction, plus two point five meters

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per second squared in the normal direction.

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I'll note that this is a vector.

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We've got direction.

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Direction is involved in the question states that we need to find the.

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The magnitude of the acceleration, so we need to take the acceleration vector and we need to find the

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magnitude of the acceleration.

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So how do we do that?

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Well, the magnitude of acceleration is the square root of the squeeze of the two components.

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We've got the square root of the tangential component plus plus the normal component squared.

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That is the square root of five squared, plus 2.5 squared.

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That is the square root of thirty one point five.

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And that gives us an answer of five point five nine meters per second squared.

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Now, this is now a scalar.

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It is not any more a vector.

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It doesn't indicate direction.

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It just tells us what is the magnitude of the acceleration.

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So if we add the two components of the acceleration together, what is the total magnitude?

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What is the size of the acceleration?

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And that is how we would answer an approach.

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Such a question.