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2) 20.2 m/s
In the first 4.4 seconds of its motion, the blue car accelerates at a rate of
So its final velocity after these 4.4 seconds is
where
u = 0 is the initial velocity (the car starts from rest)
a is the acceleration
t is the time
Substituting t = 4.4 s, we find
After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until
Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.
3) 216.2 m
The distance travelled by the car during the first 4.4 s of the motion is given by
where
u = 0 is the initial velocity
is the time
is the acceleration
Substituting,
The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is
where
is the new velocity
is the time
Substituting,
So the total distance travelled before the brakes are applied is
4)
We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is
We can find the acceleration of the car during this part by using the SUVAT equation:
where
is the final velocity (zero since the car comes to a stop)
is the velocity of the car at the moment the brakes are applied
a is the acceleration
Solving for a, we find