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Question

A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first of that distance, its acceleration is +2.25 m/s2. Through the next of that distance, its acceleration is –0.750 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?

A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +2.25 m/s2. Through the next 3/4 of that distance, its acceleration is –0.750 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?

Answer 1

It looks like you are missing a distance for each acceleration. You wrote "the first of that distance" and "the next of that distance," but I think it should be "the first n meters of that distance," where n is part of the problem statement, and the same for "the next r meters."

If you did have the distances, you would use the formula (vf)^2 = (vo)^2 + 2ax, where vf is the final velocity, vo is the initial velocity, a is acceleration, and x is the distance (n or r, depending on which part of the travel is being considered). The initial velocity is zero, but when you calculate the final velocity of the acceleration phase (which would be the maximum speed), you would then use that speed as the initial speed for the deceleration phase. Once you know the max speed and the speed at the end of the 900 m, you can find the travel times by dividing the change in velocity by the given acceleration. The change in velocity for a = 2.25 is equal to the max velocity (because the initial velocity was zero), and the change in velocity for a = -0.75 is equal to the final velocity minus the max velocity.

Answer 2

Let the dist travelled with acceleration be x1; therefore, dist travelled with de-acceleration = 900 - x1
We use the eqn v^2 = u^2 + 2 * f * h to find vel at end of first distance.
v^2 = o + 2* 2.25 * x1 = 4.5 * x1, or v = sqrt(4.5*x1)
Using the same eqn thru 2nd distance, we have
0 = v^2 - 2 * 0.75 * (900 - x1),
or 0 = 4.5 * x1 - 1.5 * (900 - x1)
Solving for x1 gives x1 = 225m for 1st dist and 675 for 2nd dist. and v = sqrt(4.5 * 225) = 31.82 max speed
From eqn v = u + f*t, 1st travel time t1 = 14.14 sec,
while 2nd travel time t2 = 42.43 sec
Total time = t1 + t2 = 56.57 sec

Answer 3

Let s1 be the distance traveled starting from rest with acceleration a1 in time t1

Let s2 be the distance traveled coming to rest with acceleration -a2 in time t2.

This can be viewed as of it has traveled a distance s2 starting from rest with acceleration a2 in time t2, viewing from the opposite direction.

S1 = 0.5 a1 t1^2 and

S2 = 0.5 a2 t2^2.

The velocities after the times t1 and t2 are the same as

v = a1 t1 = a2 t2.

t2 / t1 = (a1 /a2) = 2.75 / 0 .75 = 3.67
t2 = 3.67 t1.

S1 + S2 = 900 = 0.5 {a1 t1^2 + a2 t2^2}.

2x 900 = a1 t1^2 + a2 t2^2.

Substituting for t2 = 3.67 t1

2x 900 = 2.75 t1^2 + 0.75 x 3.67 x 3.67 t1^2.

1800 = 12.83 t1^2


t1 = 11.85 second.

t2 = 3. 67 x 11.85 = 43.47 second.

t1 +t2 = 55.32 second.

v = a1 t1 or a2 t2 = 2.75 x 11.85 = 0.75 x 43.47

=32.59m/s

Answer 4

The maximum speed is : 28.7775 m/s
and total time is 72.79 sec

Answer 5

a. 10.6265 secs.
b. 84.69 m/s