91Ó°ÊÓ

Given Data:

The acceleration of car is\(a = 1.40\;{\rm{m}}/{{\rm{s}}^2}\)

The initial speed of car is\(u = 0\;{\rm{m}}/{\rm{s}}\)

The final speed of car is\(v = 2\;{\rm{m}}/{\rm{s}}\)

The duration to stop the car is\(T = 0.800\;{\rm{s}}\)

The acceleration of a particle is the variation in the velocity of particle with time between different positions.

The duration for speeding up car is given as:

\(v = u + at\)

Here,\(t\)is the duration for speeding up car.

Substitute all the values in the above equation.

\(\begin{array}{c}2\;{\rm{m}}/{\rm{s}} = 0\;{\rm{m}}/{\rm{s}} + \left( {1.40\;{\rm{m}}/{{\rm{s}}^2}} \right)t\\t = 1.43\;{\rm{s}}\end{array}\)

Therefore, the duration for speeding up car is \(1.43\;{\rm{s}}\).

The duration for speeding up car is given as:

\(v = u - fT\)

Here,\(f\)is the deceleration of car.

Substitute all the values in the above equation.

\(\begin{array}{c}2\;{\rm{m}}/{\rm{s}} = 0\;{\rm{m}}/{\rm{s}} - f\left( {0.800\;{\rm{s}}} \right)\\f = - 2.5\;{\rm{m}}/{{\rm{s}}^2}\end{array}\)

Therefore, the deceleration of cheetah is \( - 2.5\;{\rm{m}}/{{\rm{s}}^2}\).