Q18PE
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?
Step-by-Step Solution
Verified(a) The duration for speeding up car is \({\bf{1}}{\bf{.43}}\;{\bf{s}}\).
(b) The deceleration of cheetah is \({\bf{ - 2}}{\bf{.5}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\).
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 \( - 2.5\;{\rm{m}}/{{\rm{s}}^2}\).