According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking from 62.5 mph on dry pavement, stops in 150 ft. Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up.(a) Find the magnitude of this car's acceleration while braking.(b) Find the magnitude of this car's acceleration while speeding up.(c) If its acceleration is constant while speeding up, how fast (in mph) will the car be traveling after 0.250 mi of acceleration?(d) How long does it take the car to stop while braking from 59.0 mph?

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 )  ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66²  - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

prolly a beacuse the momentuem the car has increases while going down hill