Question #226795
A truck moves from rest with a uniform acceleration of 2m/s^2 for the first 10sec, it then accelerates at a uniform rate of 1m/s^2 for another 15sec, it continues at a constant speed for 70sec an finally come to rest in 20sec by uniform retardation.
1
Expert's answer
2021-08-17T10:29:17-0400

Given:

a1=2m/s2a_1=2\:\rm m/s^2

a2=1m/s2a_2=1\:\rm m/s^2

t1=10st_1=10\:\rm s

t2=15st_2=15\:\rm s

t3=70st_3=70\:\rm s

t4=20st_4=20\:\rm s


The total distance

d=d1+d2+d3+d4d=d_1+d_2+d_3+d_4

1.

d1=a1t122=21022=100md_1=\frac{a_1t_1^2}{2}=\frac{2*10^2}{2}=100\:\rm m

2.

v2=a1t1=210=20m/sv_2=a_1t_1=2*10=20\:\rm m/s

d2=v2t2+a2t222=2015+11522=412.5md_2=v_2t_2+\frac{a_2t_2^2}{2}=20*15+\frac{1*15^2}{2}=412.5\:\rm m

3.

v3=v2+a2t2=20+115=35m/sv_3=v_2+a_2t_2=20+1*15=35\:\rm m/s

d3=v3t3=3570=2450md_3=v_3t_3=35*70=2450\:\rm m

4.

d4=v3+02t4=35+0220=350md_4=\frac{v_3+0}{2}t_4=\frac{35+0}{2}*20=350\:\rm m

Finally,

d=100+412.5+35+2450+350=3312.5md=100+412.5+35+2450+350=3312.5\:\rm m


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS