If two cars are moving, labelled car A and car B. Car A moves at 15 m/s and B at 10 m/s but car B also accelerated at 2 m/s/s. If the two both travel for ten seconds, which car will travel further?

Well for this question we can use the SUVAT equations. We'll see what values we have and what we're looking for.

So firstly we have the initial speed, often denoted usign the letter u. I'm going to write Au to mean the speed of car A.

Au=15 m/s

Bu=10 m/s

Now we also have the accelration, denoted by a. Car A doesn't change speed so its accelration is zero. Car B acceraltes by 2 m/s/s, so for every second that passes its speed increases by 2 m/s.

Aa=0

Ba=2

We also have the time taken, denoted by t. Both cars move for ten seconds so.

At=10

Bt=10

We're trying to find how far each one goes, this is denoted with the letter s.

As=?

Bs=?

With what we have we should use the SUVAT equation: s=ut+1/2at.

This equation uses the inital speed and the acceleration to find out how far an object has moved. Lets work it out for car A.

As=15 * 10+1/2 * 0 *10

As=15 * 10 = 150 m

Now for car B.

Bs=10*10 + 1/2 * 2 * 10

Bs= 100 + 10

Bs=110 m

So car A moves 150 m and car B moves 110 m, so car A moves further. 

Answered by Tom W. Physics tutor

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