[babi_gurl]

Hey hw wood I work this question out
A man runs at a velocity of 4.5m/s for 15 mn. When going up an increasingly steep hill, he
Slows down at a constant rate of 0.05m/s/s for 90 seconds and comes to a stop. How far did he run?

[Capuchin]

Well you need to split this into 2 parts, when he is decelerating and when he is not.

When he is not decelerating you need to use s = vt and when he is decelerating you need to use s = ut + 0.5at^2

get the distances and add them together!

Job's a goodun!

[babi_gurl]

so while he is not deceleration wood it be
s=vt
s=4.3 x 900
s=3870 m

and while he is decelerating
s=ut+1/2at^2
s=4.5 x 90 +0.5 x -0.05 x 90^2
s=202.5
total distance travelled = 3870 +202.5=4072.5 m

[Capuchin]

You're cleverer than you think, but you need to put 4.5 into the first part, rather than 4.3 :)

Well done! :)