Question

A certain cable car is San Francisco can stop in 10 s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d m in front of the car and slams on the brakes instantly. The car reaches the dog 7.92 s later, and the dog jumps off the track just in time. If the car travels 3.83 m beyond the position of the dog before coming to a stop, how far was the car from the dog? [Hint: You will need three equations.] m

A certain cable car is San Francisco can stop in 10 s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d m in front of the car and slams on the brakes instantly. The car reaches the dog 7.92 s later, and the dog jumps off the track just in time. If the car travels 3.83 m beyond the position of the dog before coming to a stop, how far was the car from the dog? [Hint: You will need three equations.]

Answer 1

lets say the initial speed is 'n and the relardation of the car is (-a). Time to stop Vaatat (v=0 when stops) - O = u-at - a at to u 10=u - u= loa t=19357 dye, som trios L Speed of car at t= 7.925 is = utat. sv= U- 9x7.92 =) = 4- 7.92a - ② finally the cart Stops after 3.83m. v² a = zas. (u=o) - o- (4-792a) = 2(-a) x 3.83 & (u-7.92a) = Rax3.83 2) 10a - 7.920 = 57.66a. (u= 10 a). - 2.08 a = 57.669 sovare both sides. & 4.32642 = 7.669 & a= 7.66 = 1.77mps?. 5.3264 - 10 a = 17.7m/s.

+7.928 S = ut t at² (+= 7.925 ut 17.7m/s Q- - 1.77m18?) + 1 x (-1.77)x(4.9% = 17.7x7.92 u to it.tms. Sr 84.67m Cans)