Question

A commuter backs her car out of her garage with an acceleration of\({\bf{1}}{\bf{.40}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Short answer

(a) The duration for speeding up car is\({\bf{1}}{\bf{.43}}\;{\bf{s}}\).

(b) The deceleration of cheetah is \({\bf{ - 2}}{\bf{.5}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\).

Step-by-step solution

Determination of formula for acceleration of cheetah

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 \(1.43\;{\rm{s}}\).

Determination of deceleration of car

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 \( - 2.5\;{\rm{m}}/{{\rm{s}}^2}\).