Question

2.46 The nominal stopping distances of a Nissan Sentra SE Sport Coupe are (a) 147 ft from 60.0 mi/h and (b) 264 ft from 80.0 mi/h. Determine the value of the acceleration for each case. assuming that it is constant during each event. Give your answer in units of m/s 2.47 An object at rest is subject to a constant acceleration of 2.00 m/s for 10.0 s. For the next 10.0 s there is no accelera tion. Finaily the object undergoes an acceleration of -2.00 for 10.0 s. (a) What is the final speed? (b) How far did the object go during the 30.0 s?

Answer 1

2.46

Stopping distance is given by:

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*S

S = Stopping distance

a = acceleration

U = initial Velocity

V = final Velocity = 0 m/sec

from above equation:

a = (V^2 - U^2)/(2*S)

Part A.

(1 ft = 0.3048 m)

S = 147 ft = (147 ft)*(0.3048 m/1 ft) = 44.8 m

1 hr = 3600 sec & 1 mi = 1609.34 m

U = 60 mi/hr = (60 mi/hr)*(1609.34 m/1 mi)*(1 hr/3600 sec) = 26.82 m/sec

Using these values:

a = (0^2 - 26.82^2)/(2*44.8)

a = -8.03 m/sec^2

Part B.

S = 264 ft = 264*0.3048 = 80.47 m

U = 80 mi/hr = 80*1609.34/3600 = 35.76 m/sec

Using these values:

a = (0^2 - 35.76^2)/(2*80.47)

a = -7.95 m/sec^2

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