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

Guest Jan 3, 2020

#1**+1 **

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

ElectricPavlov Jan 3, 2020