Answer:
The time it takes to stop is 13.75 seconds
Explanation:
A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;
v = u + a·t
Where;
v = The final velocity of the body
a = The acceleration of the body
t = The time of acceleration (accelerating period) of the body
u = The initial velocity of the body
The given parameters for the acceleration of the car are;
The initial velocity of the car, u = 0 m/s (a car starting from rest)
The constant acceleration of the car, a = 5.5 m/s²
The acceleration duration, t = 6 s
Therefore, we have;
The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s
The final velocity of the car after the acceleration, v = 33 m/s
When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'
The given parameters when the car slows down are;
The deceleration, -a₂ = 2.4 m/s²
The final velocity, v₂ = 0 m/s
The initial velocity, u₂ = v = 33 m/s
The time it takes to stop = t₂
-a₂ = 2.4 m/s²
∴ a₂ = -2.4 m/s²
From, v = u + a·t, we have;
v₂ = v + a₂·t₂
By plugging in the values of the variables, we have;
0 m/s = 33 m/s + (-2.4 m/s²) × t₂
∴ 2.4 m/s² × t₂ = 33 m/s
t₂ = 33 m/s/(2.4 m/s²) = 13.75 s
The time it takes to stop, t₂ = 13.75 seconds
