Let, #v_0 and v_t# be the Initial velocity and velocity after
time #t.#
Also, let #a# be the Accelearation, and #s,# the Distance.
The following well-known eqns. of motion will be used to solve the
Problem.
#(M_1) : v_t=v_0+at, and, (M_2) : v_t^2=v_0^2+2as.#
Part (a) :
#v_0=100 (km)/(hr)=(100*1000)/3600 m/(sec)=250/9 m/sec, v_t=0, t=?.#
Noting that, #a=-8 m/(sec)^2," we have, by, "(M_1), 0=250/9-8t,#
#:. t=250/(8*9)=125/36~~3.47 sec.#
Thus, the car will stop after appro. #3.47 sec.# after applying the brakes.
Part (b) :
In #(M_2),# we have, #v_t=0, v_0=250/9 m/(sec.), a=-8 m/(sec)^2, s=?#
#:. 0=(250/9)^2-16s rArr s=(250)^2/(9*16)~~434.02 m.#
Enjoy Maths.!
