Answer:
A) total time = 55.5 seconds
B) average velocity = 25.27 m/s
Explanation:
It starts from rest, so initial velocity, u = 0 m/s
We are given;
acceleration; a = 2 m/s²
Final velocity; v = 31 m/s
From Newton's first law of motion,
v = u + at
So, 31 = 0 + 2t
t = 31/2
t = 15.5 sec
We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;
(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds
b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;
S = ut + ½at²
S1 = 0 + ½(2 * 15.5²)
S1 = 240.25 m
Distance traveled in 35 sec with with velocity of 31 m/sec is;
S2 = velocity x time
S2 = 35 × 31 = 1085 m
Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.
From the first equation of motion,
a = (v - u)/t
a = (0 - 31)/5
a = -6.2 m/s²
So, distance travelled is;
S3 = ut + ½at²
S3 = (31 × 5) + ½(-6.2 × 5²)
S3 = 155 - 77.5
S3 = 77.5 m
So overall total distance = S1 + S2 + S3
Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m
Average velocity = total distance/total time
Average velocity = 1402.75/55.5 = 25.27 m/s
