A few days ago
eM-eM

# A train moving with a constant velocity of 0.5m/s is about to pass in an intersection whose distance is 8.0m.?

..cont..

At the same instant, a car with a speed of 1.5m/s & 10m away from the intersection (perpendicular to the railway), is in a hurry to beat the warning signal. If the car kept a constant acceleration of 0.25m/s^2, is it possible for the car to pass through the intersection without colliding w/ the train?..

2.) The motion of the truck is given by the equation: v=dt+ct^2, where d=50cm/s^2 & c= -5cm/s^3. Determine the..

(a) average acceleration from t=1.5s to t=5.0s

(b) instantaneous acceleration when t=3.0s

3.) the motion of a particle is described by the equation:

x=ct^2+b, where c=10cm/s^2, & b=5cm. Determine the instantaneous velocity when t=1.75s.

I really need your help guys..pls…

it’s very important and urgent..

A few days ago
³√carthagebrujah

1.) The train’s velocity and distance mean that it will arrive at the intersection in 16 seconds (8 m / .5 m/s = 16 s).

To determine if the car can make it, use the distance formula:

d = vt + 1/2at²

10m = 1.5m/s * t + 1/2 * .25 * t²

10m = 1.5t + .125t²

0 = -10 + 1.5t + .125t²

t = (-b ± √(b² – 4ac)) / 2a

t = (-1.5 ± √(2.25 + 5)) / .25

t = (-1.5 ± 2.693) / .25

t = -6 ± 10.770

t = 4.77 sec (the other answer is negative).

thus, the car will reach the intersection well before the train. Note: even if the car does not accelerate, the car reaches before the train – without acceleration, the car reaches the intersection at 10 / 1.5 = 6.667 sec. (solution)

2.) The easiest way to solve this is to remember that a = Δv/Δt (the derivative of the velocity respective to time).

So: a = d + 2ct

b: to find instantaneous acceleraton at t = 3, substitute 3 for t.

a = d + 2ct = 50 cm/s² + 2(-5 cm/s³)(3s)

a = 50 cm/s² – 30 cm/s² = 20 cm/s² (or .2 m/s) (solution)

a.) Average acceleration would mean taking a(1.5) – a(5) and dividing by 2.

a(5) = 50 cm/s² + 2(-5 cm/s³)(5s) = 50 – 50 = 0

a(1.5) = 50 cm/s² + 2(-5 cm/s³)(1.5s) = 50 – 15 = 35 cm/s²

(35 – 0) / 2 = 17.5 cm/s (solution)

3 is handled much the same way as two. v can be described as a derivative of x.

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4 years ago
balok
A practice has an engine that propels it forward, we could say at a fee of 60 mph, so being on the practice you’re additionally shifting at 60 mph. the instant your ft go away the floor your momentum for sure starts at 60 mph, in spite of the incontrovertible fact that until you have an engine up your *** or in a no ecosystem ecosystem, you’re problem to air rigidity/friction, so which you will decelerate. once you bounce in the air you’re up in the air for under a 2d, so the quantity of velocity lost is particularly measurable, so once you come down that’s fractionally off from the place you jumped. we could say you are able to bounce 50 ft in the air right now up you may lose measurable floor. even nonetheless it truly is particularly uncomplicated and entire undemanding sense they some how convince human beings otherwise because of the fact it disproves the earth rotating on the nonsensical fee we are instructed it rotates at.
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