Assume the truck is going at v0 = 25 m/s, and you're 20 m behind the truck. You decide to accelerate at a = 1.0 m/s2 to pass the
1 answer:
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(a) 25lx
(b) 11.11lx
<u>Explanation:</u>
Illuminance is inversely proportional to the square of the distance.
So,
where, k is a constant
So,
(a)
If I = 100lx and r₂ = 2r Then,
Dividing both the equation we get
When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx
(b)
If I = 100lx and r₂ = 3r Then,
Dividing equation 1 and 3 we get
When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx
Answer: a=-2.4525 m/s^2
d=s=190.3 m
Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr
Fr = 25% of weight
W=mg
W=1750*9.81
W=17167.5
Hence
Frictional force is negative as it acts in opposite direction
According to newton second law of motion
F=ma
hence
given
u= 110 km/h
u=110*1000/3600
u=30.55 m/s
to get t we know that final velocity v=0