Question

Car cruise down an expressway at 25 m/s. Engineers want to design an interchange for a deceleration of -2.0 m/s2 over 3.0 S.
What velocity will cars have at the end of the approach?
What minimum approach length will satisfy these requirements?
What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity?

Answer 1

Answer on Question 68683, Physics, Mechanics, Relativity

Question:

Car cruise down an expressway at $25\ \mathrm{m/s}$. Engineers want to design an interchange for a deceleration of $-2.0\ \mathrm{m/s^2}$ over $3.0\ \mathrm{s}$.

a) What velocity will cars have at the end of the approach?

b) What minimum approach length will satisfy these requirements?

c) What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)

Solution:

a) We can find the velocity of the car at the end of the approach from the kinematic equation:

$$v = v_0 + at,$$

here, $v_0 = 25\ \mathrm{m/s}$ is the initial velocity of the car, $v$ is the velocity of the car at the end of the approach, $a = -2.0\ \mathrm{m/s^2}$ is the deceleration of the car and $t$ is the time.

Then, we get:

$$v = v_0 + at = 25 \frac{\mathrm{m}}{\mathrm{s}} + \left(-2.0 \frac{\mathrm{m}}{\mathrm{s^2}}\right) \cdot 3.0\ \mathrm{s} = 19 \frac{\mathrm{m}}{\mathrm{s}}.$$

b) We can find the minimum approach length that will satisfy these requirements from the kinematic equation:

$$d = v_0 t + \frac{1}{2} a t^2 = 25 \frac{\mathrm{m}}{\mathrm{s}} \cdot 3.0\ \mathrm{s} + \frac{1}{2} \cdot \left(-2.0 \frac{\mathrm{m}}{\mathrm{s^2}}\right) \cdot (3.0\ \mathrm{s})^2 = 66\ \mathrm{m}.$$

c) We can find the maximum velocity of the car from the kinematic equation:

$$v^2 = v_0^2 + 2ad,$$ $$v_0 = \sqrt{v^2 - 2ad} = \sqrt{ \left(19 \frac{\mathrm{m}}{\mathrm{s}}\right)^2 - 2 \cdot \left(-8.0 \frac{\mathrm{m}}{\mathrm{s^2}}\right) \cdot 66\ \mathrm{m} } = 38 \frac{\mathrm{m}}{\mathrm{s}}.$$

Answer:

a) $v = 19 \frac{m}{s}$.

b) $d = 66 \, m$.

c) $v_0 = 38 \frac{m}{s}$.

Answer provided by https://www.AssignmentExpert.com