M1 help!!please urgent

Why t=4 or 5 they have the same speed=3?

We now are considering that she 'starts' at the top of the hill. Here, here starting velocity is $u = 3$.

We also know that the distance travelled in five seconds is equal to the distance travelled in four seconds, with 12m added on at the end.

To find the final speed, we should use $v = u + at$, but we don't know $a$. So we use $s = ut + \frac{1}{2}at^2$ to find it:

From the statement above, we can equate the two distances:

$3\times 5 + \frac{1}{2}a\times 5^2 = 12 + 3\times 4 + \frac{1}{2}a \times 4^2$

$15 + \frac{25}{2}a = 24 + 8a \Rightarrow \frac{9}{2}a = 9\Rightarrow a = 2$.

Hence $v = 3 + 2\times 5 = 13$.

Apologies for the wrong picture

We now are considering that she 'starts' at the top of the hill. Here, here starting velocity is $u = 3$.

We also know that the distance travelled in five seconds is equal to the distance travelled in four seconds, with 12m added on at the end.

To find the final speed, we should use $v = u + at$, but we don't know $a$. So we use $s = ut + \frac{1}{2}at^2$ to find it:

From the statement above, we can equate the two distances:

$3\times 5 + \frac{1}{2}a\times 5^2 = 12 + 3\times 4 + \frac{1}{2}a \times 4^2$

$15 + \frac{25}{2}a = 24 + 8a \Rightarrow \frac{9}{2}a = 9\Rightarrow a = 2$.

Hence $v = 3 + 2\times 5 = 13$.

I am a bit confused why " We also know that the distance travelled in five seconds is equal to the distance travelled in four seconds, with 12m added on at the end."
The speed is accelerating shouldn't it have different distance?
Thanks!

Well, we know that the distance travelled in the first four seconds added to the distance travelled in the fifth second must be equal to the entire distance travelled in five seconds.

As we know the distance travelled in the fifth second is 12m, it must be such that the distance travelled in the five seconds is equal to 12m added to the distance travelled in four seconds.

Well, we know that the distance travelled in the first four seconds added to the distance travelled in the fifth second must be equal to the entire distance travelled in five seconds.

As we know the distance travelled in the fifth second is 12m, it must be such that the distance travelled in the five seconds is equal to 12m added to the distance travelled in four seconds.

Perhaps if you write distance after 4 seconds is d4, distance after 5 seconds is d5, then:

d5=d4+12