It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant accelerati
on of -3.9 m/s/s, determine the speed of the Jaguar before it began to skid to a stop.
1 answer:
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Answer:
The value is
Explanation:
From the question we are told that
The magnitude of the horizontal force is
The mass of the crate is
The acceleration of the crate is
Generally the net force acting on the crate is mathematically represented as
Here is force of kinetic friction (in N) acting on the crate
So
=>
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.