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 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

Answer 1

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 motion in a circular path, 7 2 components of acceleration exists 2 normal/radial acceleration, an = 7 v velocity & radius of path Tangential acceleration, at dv dt for deceleration, ay = -dv (as spced decreases). dt Total acceleration 1) Acceleration Just aftes booking a=a= laz+ah ie, to 2 an 21.92² 8.768 mis? o 54.8 t= at dv dt = -1.370, - at = -1.37x0 = 0 2 + =) Total acceleration, a = an = 8.768 8.77 m/s2 Acceleration just aftes boaking is 8.77 m/s 2) speed at t=45 is? = -1.37 -1.374 dv dt dv a = dt love fast det Velocity at time t-o is 21.92 m/s (Just before braking v dv -- juste de 21.92 D

>> 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 -0.685x4 - 10.96 m/s velocity after 4 secs is 10.96 m/s -> acceleration at t= 45 is yan V = velocity at t=4 sec 1 an = 7 at 10.96² 54.8 = 2, 192 m/s² ay = -1.377 = -1.37x4 = -5.43 m/s ar=-ve' indicates it is opposite to velocity direction 2 Total acceleration, a = = 12.192² + (-5, 43) -> a= 5.9021 m/s 5.9 m/s Acceleration after 4 secs is 5.9 m)s -> Time to stop car stops when velocity becomes ö - (t) = 0 21.92-0.685 t = 0 2 ta = 5.656 SCC 121,92 0.685 ~ 5.66 sec .. Cal stops after t = 5.66 seconds Boaking moved distance V dr dt di = Vdt

Distance 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 - 0.6856 5.66 = 9 = 21,92x5.66 - 0.6858 3 82.67m → 5= 82.665 m Distance covered before it comes to rest is 82.67 m
I hope you understand. Distance = velocity X time. So distance covered is integrating  velocity over the time . Comment if you have any doubt.