Over a short interval the coordinate of a car in the meters isgiven by x (t) = 27t - 4.0 t3 where time t is in seconds, at the end of 1.0s the acceleration of the car is

1. 27 m/s2

2. 4.0 m/s2

3. - 4.0 m/s2

4. - 12 m/s2

5. - 24 m/s2

5. - 24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S. I unit of acceleration is m/s².

mathematically,

a = dv/dt ... Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx (t) / dt

Where,

x (t) = 27t-4.0t³ ... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx (t) / dt = 27-12t² ... Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = - 24t

a = - 24t ... Equation 4

from the question,

At the end of 1.0 s,

a = - 24 (1)

a = - 24 m/s².

Thus the acceleration = - 24 m/s²

The right option is 5. - 24 m/s²