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