Question

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because one of the experts solved it wrong...

there is no wrong in the questions!!!!

not most of them are not in the choices !!!!!

Q1) A car was moving on a circular path of radius 54.8 m with a constant speed 21.92 m/s. Suddenly the driver starts to decrease the speed at a rate 1.371 m/s?. the acceleration just after braking is Select one: a. 0.00 b. 8.77 c. 9.86 d. 9.32 e. 11.23 the speed at t4s is Select one: a. 10.96 b. 5.70 c. 3.51 d. 19.51 e. 4.16 the acceleration at t=4s is Select one: a. 5.90 b. 14.87 c. 9.56 d. 13.69 e. 1.30 time to stop Select one: a. 9.16 b. 13.12 c. 1.24 d. 5.66 e. 14.26 the braking moved distance Select one: a. 82.67 b. 26.45 c. 42.99 d. 147.15 e. 31.41

Answer 1

1) b) 8.77 m/s^2
2) a) 10.96 m/s
3) a) 5.90 m/s^2
4) d) 5.66 s
5) a) 82.67 m

just after the braks are applied, total acceleration = radianl acceleration

a = v^2/r

= 21.92^2/54.8

= 8.77 m/s^2 <<<<<<<<-------------Answer

use, a = dv/dt

dv = a*dt

integral dv = integral a*dt

vf - vi = integral -1.37*t*dt

vf - 21.92 = -1.37*t^2/2

vf = 21.92 - 1.37*t^2/2

at t = 4 s

vf = 21.92 - 1.37*4^2/2

= 10.96 m/s <<<<<<<<-------------Answer

tangential acceleration, acceleration, a = -1.37*t

= -1.37*4

= -5.48 m/s^2

radial acceleration, a_rad = v^2/r

= 10.96^2/54.8

= 2.192 m/s^2

total acceleration, a = sqrt(a_tan^2 + a_rad^2)

= sqrt(5.48^2 + 2.192^2)

= 5.90 m/s^2   <<<<<<<<-------------Answer

when the car stops, vf = 0

use, vf = 21.92 - 1.37*t^2/2

0 = 21.92 - 1.37*t^2/2

==> t = 5.66 s <<<<<<<<-------------Answer

now use,

use, v = dx/dt

dx = v*dt

integral dx = integral v*dt

= integral (21.92 - 1.37*t^2/2)*dt

= 21.92*t - (1.37/2)*t^3/3

substitute t = 5.66 s

delta_x = 21.92*5.66 - (1.37/2)*5.66^3/3

= 82.67 m <<<<<<<<-------------Answer