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On a dry road a car with good tires may be able to brake with a deceleration of 11.0 mi/h/s (= 4.92 m/s"). (a) How long does it take such a car, initially traveling at 55 mi/h (= 24.6 m/s), to come to rest? (b) How far does it travel in this time?

On a dry road a car with good tires may be able to brake with a deceleration of 11.0 mi/h/s (= 4.92 m/s"). (a) How long does it take such a car, initially traveling at 55 mi/h (= 24.6 m/s), to come to rest? (b) How far does it travel in this time?

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
6 years ago
Initial speed of car, v0 = 55 mi/h.
Final speed of car, v = 0 mi/h.
Deceleration of the car, a = -11.0 mi/h .
The negative sign of the deceleration highlights that the acceleration of the car is in a direction opposite to the direction of the initial velocity vector.
It is important to note that the decoration of the car is given in mi/h/s, therefore we first convert it in mi/h2.
The deceleration of the car in mi/h2 is:

232-2439_1.PNG
Therefore the deceleration of the car is -39,600 mi/h2.
The time t taken by the car to come at rest is given from the equation of kinematics as:
232-1788_2.PNG
The distance travelled (say x) by the car in time calculated above can be given using the equation of kinematics as:
232-817_3.PNG
It is assumed that x = 0 at t = 1, therefore the value of x0 in the above equation is zero.
Substitute the initial condition and the given values in the equation above to obtain x as:
232-1013_4.PNG
One should note that we have used time t in hours and not in seconds. The value of t is given by equation (1).
Therefore the distance x is:232-1671_5.PNG
Therefore the distance travelled by the car before coming to rest is 0.038 mi.

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