. The brakes of an unloaded lorry of mass
1000 kg will slow it down from 40 to 20 km/h
in 7.5 s. How long will they take to stop it from
a speed of 30 km/h, if it has taken on a load of
2200 kg?
(Ans. 36 s
Answers
Answer:
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Explanation:
m1=1000kg,u=40km/hr,v=20km/hr,t=7.5s
the force applied at the brakes of the unloaded lorry is F=1000(40–20)/7.5=2666.6N
Now,the truck is loaded by 2200kg
The net mass of the system is m2=2200+1000=3200kg
The time taken to stop the loaded lorry from the speed of 30km/hr with the same net force acted upon the brakes .
2666.6=3200*30/t
2.6*1000=96000/t
t=96/2.6=36.9sec
Answer:
36 s
Explanation:
mass= 1000 kg.
initial velocity = 40km/h = 40×5/18 m/s
100/9m/s.
final velocity =20 km/h = 50/9m/s.
time takin = 7.5 s
acceleration = (v-u)/t = (50/9-100/9)/7.5s
= -20/27m/s^2.
now , force = m×a
= 1000kg × -20/27 m/s^2
=-20000/27 N.
FOR SECOND CASE(after loading 2200 kg).
mass = 1000kg + 2200 kg
= 3200 kg.
we have derived force (resistance )
F = 20000/27N
acceleration = F/M
= -20000/27 N ×1/3200kg.
=25/108 m/s^2.
now we have to find time taken to stop,
we'll apply first equation ,
V= U+at
0=u+at {v=0}
t = u /a
t = 25/3÷25/108. (30 km /h = 30×5/18= 25/3 m/s).
t = 25/3 ×108/25
t = 36 s.
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