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qmaster · Aug 1, 2012, 10:22 AM

a bus is beginning to move with an accleration of 1 m/s^2. A boy who is 48m behind the bus starts running at 10m/s. after whattime will he be able to catch the bus?
a. 6s b. 8s c. 10

ebaines · Aug 1, 2012, 10:59 AM

First you need to set up the equations of motion for the bus and the boy - do you know what they are?

qmaster · Aug 1, 2012, 07:11 PM

Originally Posted by ebaines

First you need to set up the equations of motion for the bus and the boy - do you know what they are?

No I don't... plz help me

ebaines · Aug 2, 2012, 05:01 AM

You need to memorize and know how to use the following equation, which gives the position of an object as a function of time depending on its initial position x_0, its initial velocity v_0, and its acceleration a:

$<br /> x(t) = x_0 + v_0t + \frac 1 2 a t^2<br />$

For the boy to catch the bus you need to find the value for time when

$<br /> x_{boy} (t) = x_{bus} (t)<br />$

What do you think the values of x_0, v_0, and a are for the bus and the boy? Clearly you have to decide whether you want to base this on either the starting point of the bus (i.e. set x_0 for the bus = 0 and x_0 for the boy = -48) or base it on the starting point of the boy (set x_0 = 0 for the boy and x_0 =+48 for the bus). It really doesn't matter which of these you choose; you just need to be consistent. So now set up the equation and solve for t. Hope this helps.