Answer
case 1
initial speed of the car= 0 m/s
acceleration = 2 m/s²
final speed = 38 m/s
time taken to reach at the speed of 38 m/s
using equation of motion
v = u + at
38 = 0 + 2t₁
t₁ = 19 s
time for which car is moving with constant speed t₂ = 85 s
time taken to stop the vehicle t₃ = 5 s
a) Total ride time
t = t₁ + t₂ + t₃
t = 19 + 85 + 5
t = 109 s
b) distance travel in case 1
using equation of motion
v² = u² + 2 a s
38² = 2 x 2 x s₁
s₁ = 361 m
when car is moving with constant velocity
s₂ = v t
s₂ = 38 x 85
s₂ = 3230 m
when car is to stop
using equation of motion to calculate the deceleration of the car
v = u + at
0 = 38 + a x 5
a = -7.6 m/s²
distance traveled
v² = u² + 2 a s
0² = 38² - 2 x 7.6 s₃
s₃ = 95 m
total distance
s = s₁ + s₂ + s₃
s = 361 + 3230 + 95
s = 3686 m
[tex]average\ velocity = \dfrac{total\ distance}{total\ time}[/tex]
[tex]average\ velocity = \dfrac{3686}{109}[/tex]
[tex]average\ velocity =33.82\ m/s[/tex]
