Question

On a one lane road, a person driving a car at v1 = 54 mi/h suddenly notices a truck 2.4 mi in front of him. That truck is moving in the same direction at v2 = 31 mi/h. In order to avoid a collision, the person has to reduce the speed of his car to v2 during time interval Δt. The smallest magnitude of acceleration required for the car to avoid a collision is a. During this problem, assume the direction of motion of the car is the positive direction. ( I need help with b,e,g,h,i.)

(a) Enter an expression, in terms of defined quantites, for the distance \Delta x2, traveled by the truck during the time interval \Delta t.

(b) Enter an expression for the distance, \Delta x1 traveled by the car in terms of v1, v2, and a.

(c) Enter an expression for the acceleration of the car, a, in terms of v1, v2, and \Delta t.

(d) Enter an expression for \Delta x1 in terms of \Delta x2 and d, when the driver just barely avoids a collision,

(e) Enter an expression for \Delta x1 in terms of v1, v2, and \Delta t.

(f) Enter an expressionfor \Delta t in terms of v1, v2, and d.

(g) Calculate the value of \Delta t in hours.

(h) Use the expressions you entered in parts (c) and (f) and enter an expression for a in terms of v1, v2, and d.

(i) Calculate the value of a in meters per second squared.

Answer 1

Answer:

a) Δx₂ = 31*Δt

b) Δx₁ = 977.5 / a

c) a = 23 / Δt

e) Δx₁ = 42.5*Δt

g) Δt = 0.0565 h

i) a = 0.05 m/s²

Explanation:

Given

v₁ = 54 Mi/h

v₂ = 31 Mi/h

a)  We apply the formula

Δx₂ = v₂*Δt

⇒  Δx₂ = 31*Δt  (Assuming constant speed)

b) We use the formula

v₂² = v₁² - 2*a*Δx₁    ⇒   Δx₁ = (v₁² - v₂²) / (2*a)

⇒   Δx₁ = (54² - 31²) / (2*a)

⇒   Δx₁ = 977.5 / a

c) We use the equation

v₂ = v₁ - a*Δt   ⇒   a = (v₁ - v₂) / Δt

⇒   a = (54 - 31) / Δt

⇒   a = 23 / Δt

e)  We apply the formula

Δx₁ = v₁*Δt - 0.5*a*Δt²

Δx₁ = 54*Δt - 0.5*(23 / Δt)*Δt²

⇒   Δx₁ = 42.5*Δt

g) If   Δx₁ = 2.4 Mi    ⇒   2.4 = 42.5*Δt  ⇒   Δt = 0.0565 h

i) If  a = 23 / Δt  ⇒   a = 23 Mi / 0.0565 h = 407.29 Mi/h²

⇒   a = 0.05 m/s²

Answer 2

The mathematical relationship between parameters of an object in motion, are given by the kinematic equations of motion

The responses to the questions are:

(a) Δx₂ = v₂ × Δt

(b) $\Delta x_1 = \mathbf{\dfrac{v_2^2 - v_1^2}{2 \times a}}$

(c) $a = \mathbf{ \dfrac{v_1 - v_2 }{\Delta t}}$

(d) Δx₁ = Δx₂ + d

(e) $\Delta x_1 = \mathbf{\dfrac{(v_1 + v_2) }{ 2 } \times \Delta t}$

(f) $\Delta t = \mathbf{\dfrac{2 \cdot d}{v_1 - v_2}}$

(g) ≈ 0.21 hours

(h) $The \ acceleration \ a = \mathbf{\dfrac{(v_1 - v_2)^2}{2 \cdot d}}$

(i) a ≈ 0.014 m/s²

The reason the above values are correct are as follows:

The given parameters are;

The velocity of the car, v₁ = 54 mi/h

Distance of the truck ahead, d = 2.4 mi

Direction of the truck ahead = The same direction

Speed of the truck, v₂ = 31 mi/h

Speed to which the car reduces to = v₂

Time it takes the car to reduce speed to v₂ = Δt

Minimum acceleration magnitude of the car = a

The cars direction of motion = The positive direction

(a) Required:

To write an expression, in terms of defined quantities, for the distance Δx₂ traveled by the truck in time Δt

Solution;

Given that the truck is traveling with constant velocity, we have;

Δx₂ = v₂ × Δt

(b) Required:

An expression for Δx₁, in terms of v₁, v₂, and a

Solution;

The appropriate kinematic of motion to use is v² = u² - 2·a·s

Where:

v = The final velocity

u = The initial velocity of the car

a = The acceleration

s = The distance travelled

Therefore, we have;

v₂² = v₁² - 2 × a × Δx₁

Therefore, the expression for Δx₁, is presented as follows;

$\Delta x_1 = \mathbf{\dfrac{v_2^2 - v_1^2}{2 \times a}}$

(c) Required:

To enter an expression for the acceleration of the car, a, in terms of v₁, v₂, and Δt

Solution;

The appropriate kinematic equation of motion to use is v = u + a·t

Therefore;

v₂ = v₁ - a·Δt

Which gives;

$a = \mathbf{ \dfrac{v_1 - v_2 }{\Delta t}}$

(d) Required:

To express Δx₁, in terms of Δx₂, and d, when collision is barely avoided by the driver

Solution:

Δx₁ = Δx₂ + d

(e) Required:

To write an expression for Δx₁ in terms of v₁, v₂, and Δt

Solution:

The appropriate kinematic equation of motion to adapt is s = u·t - (1/2)·a·t²

v₂ = v₁ - a·t

$a = \dfrac{v_1 - v_2 }{\Delta t}$

Therefore, we have;

$\Delta x_1 = v_1 \cdot \Delta t - \left(\dfrac{1}{2} \right) \times \dfrac{v_1 - v_2 }{ \Delta t } \times \left(\Delta t \right)^2 = \dfrac{(v_1 + v_2) }{ 2 } \times \Delta t$

$\Delta x_1 = \mathbf{\dfrac{(v_1 + v_2) }{ 2 } \times \Delta t}$

(f) Required:

To write an expression for Δt in terms of v₁, v₂ and d

solution:

$\Delta x_1 = \dfrac{(v_1 + v_2) }{ 2 } \times \Delta t$

$\Delta x_2 + d = \dfrac{(v_1 + v_2) }{ 2 } \times \Delta t$

$v_2 \times \Delta t + d = \dfrac{(v_1 + v_2) }{ 2 } \times \Delta t$

$d = \dfrac{(v_1 + v_2) }{ 2 } \times \Delta t - v_2 \times \Delta t$

Using a graphing calculator, we have;

$\Delta t = \mathbf{\dfrac{2 \cdot d}{v_1 - v_2}}$

(g) Required:

To calculate Δt in hours

Solution:

From $\Delta t = \dfrac{2 \cdot d}{v_1 - v_2}$, we have;

$\Delta t = \dfrac{2 \times 2.4 \, mi}{54 \ mi/h - 31 \, mi/h} =\dfrac{24}{115} h$

Δt = 24/115 hours ≈ 0.21 hours

(h) Required:

To use the expression you entered in part (c) and (f) and enter an for a in terms of v₁, v₂ and d

solution:

From (c), we have;

$\Delta t = \dfrac{2 \cdot d}{v_1 - v_2}$

From (f) we have

$a = \dfrac{v_1 - v_2 }{\Delta t}$

We get;

$a = \dfrac{v_1 - v_2 }{ \dfrac{2 \cdot d}{v_1 - v_2}} = \dfrac{(v_1 - v_2)^2}{2 \cdot d}$

$The \ acceleration \ a = \mathbf{\dfrac{(v_1 - v_2)^2}{2 \cdot d}}$

(i) Required:

To calculate the value of a in meters per second squared

Solution:

From $a = \dfrac{(v_1 - v_2)^2}{2 \cdot d}$, we get;

$a = \dfrac{(54 - 31)^2}{2 \times 2.4} =\dfrac{2,645}{24} \approx 110.21$

The acceleration of the car ≈ 125.95 mi/h²

1 mile = 1609.344 meters

125.95 mi/h² = 125.95 mi/h² × 1,609.344 m/mi × (hr ²) / (3,600 s)² ≈ 0.014 m/s²

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