Hello!

When the acceleration `a` is constant, the speed `V` is equal to `V(t)=V_0+at,` where `V_0` is the initial speed. The distance traveled is `D(t)=V_0 t+(a t^2)/2.`

**1** ) The final speed is `V_1=V_0t+at=0+5*8=40(m/s).` The distance of accelerated motion is `0+(5*8^2)/2=160(m),` the distance of constant speed motion is `V_1*(12-8)=40*4=160(m).` So...

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Hello!

When the acceleration `a` is constant, the speed `V` is equal to `V(t)=V_0+at,` where `V_0` is the initial speed. The distance traveled is `D(t)=V_0 t+(a t^2)/2.`

**1**) The final speed is `V_1=V_0t+at=0+5*8=40(m/s).` The distance of accelerated motion is `0+(5*8^2)/2=160(m),` the distance of constant speed motion is `V_1*(12-8)=40*4=160(m).` So the total distance is **320m**.

**4**) (The same formulas): probably at both periods the acceleration is constant and is the same.

For the first period `D_1=20=0+(a*2^2)/2,` so `a=10(m/s^2),` and `V_1=0+a*2=20(m/s).`

The data for the second period give us nothing new, although there is no contradiction: `D_2=160=V_1*4+(10*4^2)/2=80+80` is true.

So after `7` seconds the speed will be `a*7=70(m/s).`

**2**) This is a different but simple story. The average speed is by definition (the total distance traveled) / (total time spent). In our case it is

`(2 AB)/((AB)/30 + (AB)/20) = 2/(1/30+1/20)=(2*20*30)/(20+30)=24(m/s).`

Here AB denotes the distance between A and B.