Answer:
(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
