physics task

You get the maximum velocity difference of you choose 95 and 45 km/h and if you divide it with the shortest time, you get the maximum acceleration.

You get the minimum velocity difference of you choose 85 and 55 km/h and if you divide it with the longest time, you get the minimum acceleration.

You have the right idea of how to approach a problem like this, you've just gotten some details wrong.

First off, I think that this problem becomes easier if we instead use the formula

$a=\dfrac{\Delta v}{\Delta t}$

We first want to figure out what makes $a$ as large as possible. A quotient is at its largest when the numerator is large and the denominator is small. We therefore want $\Delta v$ to be large and $\Delta t$ to be small. To make $\Delta v$ large, we want the difference between $v_1$ and $v_2$ to be as large as possible. This occurs when $v_1=95\ \text{km/h}$ and $v_2=45\ \text{km/h}$ such that $\Delta v=50\ \text{km/h}$. We want $\Delta t$ to be as small as possible, which means we take $\Delta t=0,45\ \text{s}$. What do we then get as our value for $a_\text{max}$?

Can you carry out a similar approach for $a_\text{min}$?

I tried but I did not get the right answer. 

a(max) = delta v / delta t 

the velocity here should be 95km/h which is approximately 26.38 m/S . 
I have to divide it with 0.45 s in order to get the maximum acceleration. 

a(max) = v(max)/t(min)= 26.38/0.45= 51.95m/s^2 . 

a(min) = v(min)/t(max)

v(min) =(50km/h-0.5 km/h)=13.75m/s 

t(max)=0.55 s 

a(min) = 13.75/0.55=25m/s ^2

~~~~
(25+51.95)/2=38.477m/s^2 

delta a = 51.95-38.477=13.477

38.5 +- 13.5 m/s^2 . I can’t go further with my calculations......I’ve stumbled here😩

I have solved the question. Thank you for your help!!