Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut + a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = a
= x 14 x
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
Answer:
33.6371 m
Explanation:
t = Time taken
u = Initial velocity = 20.3 m/s
v = Final velocity
s = Displacement
a = Acceleration = -7 m/s²
Distance traveled in the 0.207 seconds
Distance = Speed × Time
⇒Distance = 20.3×0.207 = 4.2021 m
Equation of motion
Distance traveled by the car while braking is 29.435 m
Total distance measured from the point where the driver first notices the red light is 29.435+4.2021 = 33.6371 m
Think of a wedge as something you put in between objects, so it is a separates objects
D if the blue car started higher it would have more energy but since the red car is lower it is going faster because it’s going down a hill