Acar starts from rest and travels to the north for 6.00 s with a uniform acceleration of 2.50 m/s2. the driver then applies the brakes, causing a uniform acceleration of –2.00 m/s2. if the brakes are applied for 4.00 s, (i) how fast is the car going at the end of the braking period, and [7 m/s] [2 marks](ii) how far has the car gone? [89 m] [3 marks]

To calculate the final velocity, we use Newton's first equation of linear motion:v=u+at
Where v is final velocity
u is initial velocity
a is the average acceleration
t is the time taken during acceleration.
Therefore,
v=0+2.5m/s²*6.00s
=15m/s
Decelerating from 15m/s;
v=15m/s+(-2m/s²×4.0s)
=3m/s To get the distance it travelled, we use v²=u²+2as During acceleration, the distance travelled is calculated as below. 15²=0+2×2.5S 225=5S S=45meters During decellaration, displacement is calculated as below, 3²=15²+(2×4S) 9=225+8S 8S=216 S=27meters Total displacement=45m+27m =72 meters.

The answer is b.the gaseous state is electrically positive, while the plasma state is electrically negative.