Let, v_0 and v_tv0andvt be the Initial velocity and velocity after
time t.t.
Also, let aa be the Accelearation, and s,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.(M1):vt=v0+at,and,(M2):v2t=v20+2as.
Part (a) :
v_0=100 (km)/(hr)=(100*1000)/3600 m/(sec)=250/9 m/sec, v_t=0, t=?.v0=100kmhr=100⋅10003600msec=2509msec,vt=0,t=?.
Noting that, a=-8 m/(sec)^2," we have, by, "(M_1), 0=250/9-8t,a=−8m(sec)2, we have, by, (M1),0=2509−8t,
:. 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.!
