An electric train moves from rest with a uniform acceleration of 1.5 for the first 10s and continues accelerating at 0.5 for a further 20s. It continues at a constant speed for 90s and finally takes 30s to decelerate uniformly to rest.
Using the equations of motion calculate the total distance travelled
a1=1.5,a_1=1.5,a1=1.5, t1=10,t_1=10,t1=10, a2=0.5,a_2=0.5,a2=0.5, t2=20,t_2=20,t2=20, t3=90,t_3=90,t3=90, t4=30.t_4=30.t4=30.
s1=a1t122=75 m.s_1=\frac{a_1 t_1^2}{2}=75~m.s1=2a1t12=75 m.
v1=a1t1=15 ms,v_1=a_1t_1=15~\frac ms,v1=a1t1=15 sm,
s2=v1t2+a2t222=400 m.s_2=v_1t_2+\frac{a_2t_2^2}{2}=400~m.s2=v1t2+2a2t22=400 m.
v2=v1+a2t2=25 ms,v_2=v_1+a_2t_2=25~\frac ms,v2=v1+a2t2=25 sm,
s3=v2t3=225 m.s_3=v_2t_3=225~m.s3=v2t3=225 m.
0=v2−a3t4,0=v_2-a_3t_4,0=v2−a3t4,
a3=v2t4,a_3=\frac{v_2}{t_4},a3=t4v2,
s4=v2t4−a3t422=v2t42=375 m.s_4=v_2t_4-\frac{a_3t_4^2}{2}=\frac{v_2t_4}{2}=375~m.s4=v2t4−2a3t42=2v2t4=375 m.
s=s1+s2+s3+s4=1075 m.s=s_1+s_2+s_3+s_4=1075~m.s=s1+s2+s3+s4=1075 m.
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