Asked by veverly
A car is traveling in a straight line at a constant speed of 20.0 m/s. Suddenly, the driver steps on the brake which exerted a constant stopping force of 15.0 N, bringing the car to a stop at 62.5 m.
a. Draw a diagram showing the direction of the stopping force? (5 pts)
b. How long would it take for the car to come to a stop? (5 pts)
c. What is the car’s mass? (5 pts)
a. Draw a diagram showing the direction of the stopping force? (5 pts)
b. How long would it take for the car to come to a stop? (5 pts)
c. What is the car’s mass? (5 pts)
Answers
Answered by
Anonymous
v = Vi + a t
0 = 20 + a t
so
a t = - 20
x = Xi + Vi t + (a/2) t^2
62 5 = 20 t + (a/2) t^2
butt t = -20/a
62.5 = 20 (-20/a) +(a/2)(400/a^2 ) = -200/a
a = -200/62.5
t = -20/a
You can take it from there. Remember m = F/a
0 = 20 + a t
so
a t = - 20
x = Xi + Vi t + (a/2) t^2
62 5 = 20 t + (a/2) t^2
butt t = -20/a
62.5 = 20 (-20/a) +(a/2)(400/a^2 ) = -200/a
a = -200/62.5
t = -20/a
You can take it from there. Remember m = F/a
Answered by
oobleck
(b) a = 20/t, so
20t - 1/2 (20/t) t^2 = 62.5
t = 6.25 s
(c) F = ma, so
m = 15/(20/6.25) = 4.6875 kg
pretty small car!
20t - 1/2 (20/t) t^2 = 62.5
t = 6.25 s
(c) F = ma, so
m = 15/(20/6.25) = 4.6875 kg
pretty small car!
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