Distance run by the train at the moment t:
x = a * t^2 / 2
Distance run by the man, D
D = 9 m + distance run by the train
D = 9m + a(t^2) / 2
Velocity of the man, V = D/t => D = V*t
Equal D:
V*t = 9m + a(t^2) / 2
Now, you have to make an additional assumption. It is that when the man reaches the traing their velocities are the same. This if because, if the train is faster, the man will not reach it, and if the train is slower then he will pass the train (which is an unnecessary waste of effort from the man)
So, the best contition is that the speed of the train equals the speed of the man.
The speed of the train follows the equation of the uniformly accelerated motion: V = a.t
So from substitute that V in the equation V*t = 9m + a(t^2)/2
=> (at) t = 9m + a(t^2) / 2
=> a(t^2) - a(t^2) / 2 = 9m
=> a(t^2) / 2 = 9m
=> (t^2) = 9m * 2 / a = 9m * 2 / ( 2 m/s^2) = 9 s^2
=> t = 3 s
=> D = 9m + a(t^2) / 2 = 9 m + 2(m/s^2) (3s)^2 / 2 = 9m + 9 m = 18 m
=> V = D / t = 18 m / 3s = 6m/s
Answer:
The man ran 18 m.
He got into the train after 3 s
The full speed is 6 m/s
Answer:
Explanation:
Given that,
Mass, m = 27 grams
Volume of the substance, V = 15 cm³
We need to find the density of the substance. We know that, the density of an object is given by mass per unit volume. So,
So, the density of the substance is equal to 
.
<span>The person is dragging
with a force of 58 lbs at an angle of 27 degrees relating to the ground. We
want to use cosine function to look for the horizontal force component. And
then we can compute for W = (Horizontal Force) x (Distance). We want the
horizontal force component since that is the component that is parallel to the
direction the cart is moving. </span><span>
(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal
force component.)
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>
Answer:
C)Both A and B
Explanation:
It is best practice to replace the worn out tie rod with Pairs instead of a single tie rod. It is not absolutely necessary, but many technicians insist on replacing both even if only one is worn out. This is because both have more or less the same load and sooner or later the other one is going to fail too. That's why technician A said that outer tie rod ends should be replaced in pairs, even if only one is worn. Technician B said that inner tie rod ends should be replaced in pairs, even if only one is worn. Both the technicians are correct
