a) accelerated motion what we know: initial velocity vi = 0, time t = 3s, acceleration a = 40m/s²; the formula to find the space is: s = vi + 1/2·a·t² all the units of measurement are fine, therefore: s = 0 + 1/2·4·9 = 18m
b) constant motion we know t = 2s and that the velocity here is equal to the final velocity (vf) of part a), which we need to calculate: vf = vi + a·t which brings vf = 0 + 4·3 = 12m/s; we can know calculate the total space with the formula s = vf·t = 12·2 = 24m
c) decelerated motion we know the acceleration a = -3m/s² (minus because the velocity is decresing) and the initial velocity which is equals to the velocity of part b), therefore vi = 12m/s we need to calculate first the time taken to bring the car to stop, we can use the formula t = v/a = 12/3 = 4s we can know calculate the space through the formula s = vi·t + 1/2·a·t² = 12·4 + 1/2·(-3)·(4²) = 48 - 24 = 24m
We can now sum up the three spaces found to get the total distance between the stop signs: s = sa + sb + sc = 18+24+24 = 66m
