SOMEONE HELP ME LIKE ASAP THIS IS DUE AT 1:50pm
2 answers:
You might be interested in
Answer:
The rate at which the distance between the two cars is changing is;
15.53 ft/sec.
Step-by-step explanation:
To solve the question, we note that
Position of car A 4 s after start of motion, w = 18 ft west,
Position of car B 4 s after start of motion, n = 27 ft north
Therefore
The distance between the two cars at the 4 s instance is
d² = w² + n²
d² = 18² + 27² = 1053 ft² and
d = 32.45 ft
The rate at which the distance between the two cars is changing is given by;
Differentiating both sides of the equation, d² = w² + n², with respect to t as follows.
It is given that the speeds of car A and car B at the 4 second instant are 7 ft/sec and 14 ft/sec, respectively
That is;
and
Substituting the values of speed in the equation of rate of change gives
The rate at which the distance between the two cars is changing = dd/dt = 15.53 ft/sec.