hello i need help figuring out this question

Question

A rapid transit trolley moves with a constant acceleration and covers the distance between two points 300 feet apart in 8 seconds. Its velocity as it passes the second point is 50 feet per second.

a)  What is its acceleration?

b)  What is the velocity of the trolley as it passes the first point?

Answer 1

Let s(t) := the trolley's distance from the first point. Let's set s(0) = 0 (i.e., frame the problem so the trolley is at the first point at time 0).

We are told that s(8) = 300 and s'(8) = 50

Then, because acceleration (the second derivative) is constant and s(0) = 0,

s(t) must be of the form at² + bt

300 = s(8) = a*8² + b*8 = 64a + 8b

50 = s'(8) = 2a*8 + b = 16a + b

Multiply the second equation by 4 and subtract the result from the first equation to get

100 = 4b, therefore b = 25

Substitute b = 25 into either of the original equations. I'll do the first one:

300 = 64a + 8*25, therefore a = 100/64 = 25/16

Answer to Part A) 25/16

For Part B) We're asked to evaluate s'(0). We now know

s(t) = 25t²/16 + 25t, so s'(t) = 25t/8 + 25

Answer to Part B) is s'(0), which equals 25

Answer 2

One formula that you could use here is:

s(t) = s₀ + ½ (v₀ + v_f) * t

where:

t = time

s(t) = displacement in terms of time

s₀ = the displacement when t = 0

s_f = the final displacement (when time = 8 seconds)

v₀ = the velocity when t = 0

v_f = the final velocity (when time = 8 seconds)

a = the constant acceleration

Given:

s₀ = 0

And when t = 8 seconds:

s(8) = s_f = 300 ft

v(8) = v_f = 50 sec

s(t) = s₀ + ½ (v₀ + v_f) * t

s(8) = 0 + ½ (v₀ + 50) * 8 = 300

(v₀ + 50) * 4 = 300

v₀ + 50 = 75

v₀ = 25 ft/sec

a = (50 ft/sec - 25 ft/sec) / 8 sec

a = 25/8 ft / sec²

a) What is its acceleration?

>>> Its constant acceleration is 25/8 ft / sec².

b) What is the velocity of the trolley as it passes the first point?

>>> Its velocity as it passes the first point is 25 ft/sec.

Answer 3

distance = acceleration times time squared + velocity times time + initial distance

d = at^2 +vot

vo = velocity at first point

300 = (a/2)(8)^2 + vo(8)

300 = 64a +8vo

50 = at +vo

vo = 50 -a(8) = 50-8a

300 = 64a/2 + 8(50-8a) = 32a+400-64a

32a=100

a = 100/32=3,125 ft/sec^2

vo=50-8a=50-8(100/32)=100/4= 25 ft/sec

initial velocity at point 1 was 25 feet per second

the constant acceleration was 3 1/8 feet per second per second

after 8 seconds, velocity increased by 8a = 8(100/32)= 25 additional feet per second

25+25 = 50 feet per second 8 seconds later at point 2

acceleration = the derivative of velocity

velocity = derivative of distance

The trolley doubled its speed in 8 seconds, from 25 to 50 feet per second

jerk is the derivative of acceleration = 0 when acceleration is constant

the general formula is distance = (a/2)t^2 + vt

some people forget to put the 1/2 with a, and just put a. That throws off your answer.

For example, when gravity is the acceleration, the formula is height = -16t^2 + vt + ho

It's only when you take the derivative that you get -32t +v where -32 ft/sec/sec is the deceleration due to gravity.

Answer 4

This is really a physics question, but I think I can help.

The equations you need are:

v(t) = v(0) + at

and s(t) = s(0) = v(0)t = .5at²

where the symbols have the following meaning

s(t) is the distance traveled in t seconds and s(0) the initial distance is 0,

v(t) is the velocity at time t and a is the constant acceleration.

Now plug in the values:

from the distance equation: 300= 8v(0) + 32a

and from the velocity equation: 50 = v(0) + 8a

This is a simultaneous set in the variable v(0) and a which you can solve by dividing the

first equation by 4 and subtracting, which will get you v(0)=25.

You can then get a.

A tutor who is more knowledgeable about motion than I am may be able to help better.