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# Acceleration and deceleration of a car

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a. A car drives up at a speed of 20 m/s and brakes. Show that the deceleration of the car is equal to 2,0 m/s^2 and that the braking distance is 100 meters.

b. A car drives up at a speed of 254 km/h and brakes maximal. The maximal deceleration is 5,0 m/s^2. Show that the braking distance is equal to 23 m.

c. A car drives up at a speed of 54 km/h on an acceleration lane and speeds up till his speed is 126 km/h. The speeding up takes 8,0s. Show that the acceleration is equal to 2,5 m/s^2

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Jan 3, 2020

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Acceleration is change in velocity divided by change in time.

vf = vo + at       velocity final = velocity original + acceleration * time

0 = 20 m/s + (-2 m/s^2) t       solve for t = 10 sec

xf = x0 + vo t + 1/2 a t^2

xf = 0 + 20 (10) + 1/2 (-2)(10^2)

xf = 200 + (-100)

xf = final position = 100 m                     Correct !

Units must match !      (this is fast 254 km/hr = 159 miles per hour !)

254 km/hr  x 1000m/km  x 1 hr/3600s = 70.556 m/s  = vo

vf = 0 = vo + (a) t =

0 = 70.556 m/s    + (-5 m/s^2)* t      solve for t = 14.11 sec

xf = x0 + vo t + 1/2 (a) t^2

xf = 0 + 70.556  (14.11) + 1/2 (-5) (14.11)^2  =  497.9 m      It cannot stop in 23 m (unless it hits a wall !)

change in velocity / change in time =   remember to keep you 'units' matched !

change in velocity   is    126- 54  km/hr   Thi s equals   72 km/hr x 1000m/km x 1hr/3600 s =  20 m/s

now     change in velocity / change in time =  20 m/s / 8 s = 2.5 m/s^2

Jan 3, 2020