Question #239815

A car from rest starts and accelerates 6 𝑚/𝑠 2 for 5 𝑠, after which it travels with a constant velocity for 9 𝑠. The brakes are then applied so that it decelerates at 4 𝑚/𝑠 2 . Find the total distance traveled by car.


1
Expert's answer
2021-09-24T09:27:07-0400

Given:

v1=0m/s,a1=6m/s2,t1=5s;v_1=0\:{\rm m/s},\quad a_1=6\:{\rm m/s^2},\quad t_1=5\:\rm s;

a2=0m/s2,t2=9s;a_2=0\:{\rm m/s^2},\quad t_2=9\:\rm s;

a3=4m/s2,v4=0m/s.a_3=-4\:{\rm m/s^2},\quad v_4=0\:\rm m/s.


The total distance

d=d1+d2+d3d=d_1+d_2+d_3

d=(a1t12/2)+(a1t1t2)+(v42(a1t1)22a3)d=(a_1t_1^2/2)+(a_1t_1*t_2)+\left(\frac{v_4^2-(a_1t_1)^2}{2a_3}\right)

=(652/2)+(659)+(02(65)22(4))=457.5m=(6*5^2/2)+(6*5*9)+\left(\frac{0^2-(6*5)^2}{2*(-4)}\right)=457.5\:\rm m


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