# 1D Kinematics question, need help!

A sprinter running a 100 meter race starts at rest, accelerates at constant acceleration
with magnitude A for 2 seconds, and then runs at constant speed until the end.

a) Find the position (relative to the start position) and speed of the runner at the end of
the 2 seconds in terms of A. ANS: 2A , Vx=2A

b) Assume that the runner takes a total of 10 seconds to run the 100 meters. Find the
value of the acceleration A. You can leave your answer in terms of a fraction but clearly
indicate the units. ANS: 100/18

.....struggling with b) !!!!!!!!!!!!!!!!

jgens
Gold Member
You need to show us some work before we can help.

Relevant equations
all kinematic eq'n with x, v, a in it

What I know so far:

I got that x(2)=2A and v(2)=2A

when it comes to b) i have tried various inputs...
Which way do I approach this??? Do approach it from x=100 and t=10, do I incorporate x=2A and v=2A.

The closest I've gotten is using x=100-2A, t=8, v=2A...

Consider the 100m broken down into two parts
- the part with constant acceleration from part a). You were asked to get the displacement of this in terms of A in part a). This will help you with that.
- and consider the second part where velocity is constant. what formulae can you use?

according to your post, it can be broken down into two times

t=2 and t=8
at t=2 , what I know is x=2A, v=2A, a=A

at t=10, what I know is x=100, v=10 (because v=d/t=100/10, is this relavant?)

there are two eq'ns that come to mind

x(t)= Xi + Vit + 0.5at^2

Vf^2= Vi^2+2a(Xf - Xi)

For the second part of the race, the eight seconds at constant velocity, just think displacement = velocity X time.
Your v=d/t = 100/10 isn't correct because velocity wasn't constant for the entire 100m.
For the first two seconds, acceleration was constant, and velocity was increasing (at a constant rate= acceleration). For the remaining eight seconds then, velocity is constant, so you can use v = d/t here.