A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s. If the car after this acceleration slows down uniformly at a rate of 2.4m/s, how long does it take to stop​

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

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explanation: