Question #81be9

1 Answer
Jan 13, 2017

#"160 m s"^(-1)#


The cool thing to notice here is that you can solve this problem without using any equations; all you have to do here is to use the definition of acceleration.

As you know, an object's acceleration tells you how the velocity of said object is changing in time.

#"acceleration" = "change in velocity"/"chaneg in time" = (Deltav)/(Deltat)#

More specifically, acceleration represents a measure of how fast the velocity of an object is changing per second.

In your case, the plane has an acceleration of #"4 m s"^(-2)#. Notice that you can write this as

#"4 m s"^(-2) = "4 m s"^(-1) * "s"^(-1) = "4 m/s"/"1 s" color(white)(color(blue)( -> " change in velocity")/(color(blue)(->" per second"))#

This is equivalent to saying that the velocity of the place increases by #"4 m s"^(-1)# with every passing second. Since you know that the plane is accelerating for #"40 s"#, you can say that its velocity will increase by

#40 color(red)(cancel(color(black)("s"))) * "4 m s"^(-1)/(1color(red)(cancel(color(black)("s")))) = "160 m s"^(-1)#

If you assume that the plane is starting from rest, i.e. with an initial velocity equal to #"0 m s"^(-1)#, you can say that its velocity at takeoff will be

#v_"takeoff" = "0 m s"^(-1) + overbrace("160 m s"^(-1))^(color(blue)("increase due to acceleration")) = color(darkgreen)(ul(color(black)("160 m s"^(-1))))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for your data.