Ajogger accelerates from rest to 3.0 m/s in 2.0 s. a car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) find the acceleration (magnitude only) of the jogger. (b) determine the acceleration (magnitude only) of the car. (c) does the car travel farther than the jogger during the 2.0 s? if so, how much farther?

(a)  a₁:  jogger  acceleration= 1.5 m/s²

(b)  a₂:  car  acceleration = 1.5 m/s²

(b)  d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁:  jogger displacement   

t₁ :  jogger time

v₀₁:  jogger initial speed

vf₁:  jogger  final speed

a₁:  jogger  acceleration

d₂: car displacement   

t₂ : car  time

v₀₂: car  initial speed

vf₂:  car  final speed

a₂:  car  acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

(c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

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explanation:

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