To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 58.6 m when its initial speed is 79.2 km/h, and 24.2 m when its initial speed is 47.8 km/h. What are (a) your reaction time and (b) the magnitude of the deceleration

a.) 2.66 seconds for 79.2 km/h and 1.82 seconds for 47.8 km/h

b.)-4.13 m/s2 for 79.2 km/h and -3.63 m/s2 for 47.8 km/h

Explanation:

Now for (a)

time = velocity/distance

for velocity = 79.2 km/h and distance 0.0586 km

t = (0.0586/79.2)*3600

t = 2.66 seconds (Please note that multiplication with 3600 is to convert hours into seconds)

for velocity 47.8 km/h and distance 0.0242 km

t = (0.0242/47.8)*3600

t = 1.82 seconds

Now for (b)

2as = vf2-vi2

so,

2a(58.6) = -(79.2*1000/3600)^2 (Please note that multiplication and division with 1000 and 3600 respectively is to convert speed unit from km/h to m/s)

a = -4.13 m/s2 for 58.6 m

and

2a(24.2) = -(47.8*1000/3600)^2

a = -3.63 m/s2 for 24.2 m.

Please note that "-" sign express the deceleration.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

answer; how far out of place which the object finished relative to the starting location;