A few days ago

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


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..

lots of thanks in advance..(^_^)

m just asking for your help..i find it hard to answer..and the deadline of this is already tomorrow..need your help..even just 1 or 2 answers that you could give from these questions are highly appreciated..pls help me..thanks..

Top 2 Answers
A few days ago

Favorite Answer

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

Use the quadratic equation:

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.


4 years ago
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