1 answer

I need right Answer ASAP, please. Q1) A car was moving on a circular path of...

Question:

I need right Answer ASAP, please.
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 decre
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
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 1.370 m/s2. 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 t-4 s 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

Answers

can moving in a circular path of fadius 54.8m with constant Speed 21.92 m/s an deceleration is given by 1.370 ms? v2 at - For>> V-21.92 = -1.376 2 > velocity VLt) = 21.92 -1.37% velocity after brake is applied, VLL) = 21.92 -0.685 t V[4) = 21.92 -0.6Distance covered in time t= 5,66 seconds is 5.66 distance is di - S (21.92 -0.685t”) at 0 t=0 5.66 5 = 21.920 21.926 - 0.6856I hope you understand. Distance = velocity X time. So distance covered is integrating  velocity over the time . Comment if you have any doubt.

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