Hello!

The acceleration is probably considered constant (uniform). Denote it as `a(m/s^2).`

Then during each second Jill's speed `V` becomes more by `a m/s.` Therefore for any time `t` after the start the speed is `V(t) = V_0 + a t,` where `V_0 = 20 m/s` is the initial speed.

It is given...

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Hello!

The acceleration is probably considered constant (uniform). Denote it as `a(m/s^2).`

Then during each second Jill's speed `V` becomes more by `a m/s.` Therefore for any time `t` after the start the speed is `V(t) = V_0 + a t,` where `V_0 = 20 m/s` is the initial speed.

It is given that for the final moment `t_1 = 10 s` the speed is `V_1 = 40 m/s.` Therefore `V_1 = V(t_1) = V_0 + a t_1,` and we can easily find `a = (V_1-V_0)/t_1.`

In numbers it is equal to `(40-20)/10 = 2 (m/s^2).` This is the answer.

That said, `40 m/s` and even `20 m/s` is too high speed for running, probably Jill used some equipment, for example a bicycle. Also note that the same speeds could be reached within non-constant acceleration, but there are infinitely many non-constant functions `a(t)` which satisfy the given conditions.