Problem:
"You are running down a football field (let’s call this the x-direction) with a constant velocity of 4.5 m/s when a player from the opposing team runs towards you, hitting you in a direction perpendicular to your motion (let’s call this the y-direction). The hit causes you to accelerate for a period of 18 s in the direction y-direction. If you move equal distances in the x and y directions during this time, what is the magnitude of the acceleration?"
The answer provided from the text is a = 0.5 m/s^2
I'm not quite sure how to set up this problem and what equations I should use.
What I know:
- t = 18 s
- Vix (initial velocity in the x-direction) = 4.5 m/s
- The distance traveled in the x-direction is equal to the distance traveled in the y-direction, which means the football player traveled at a 45 degree angle.
I feel like I still don't have enough information. Is there something that I am missing?
Thanks!
Using $s=ut+\frac 12at^2$, the distance in the $y$ direction travelled in 18 seconds is $0+\frac 12 a\times 18^2$.
There is no change in speed along the $x$ axis, so the distance travelled along the $x$ axis in this time is $4.5\times 18$. Making these distances equal gives $a=0.5$.