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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

Respuesta :

Answer:

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.

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