Question

A car accelerates from rest at 1.0 m/s^2 for 20s along a straight roa, it then moves at a constant speed for half an hour, it then deccelerates uniformly to a stop in 30s. Find the total distance covered by the car.

Answer 1

first leg:
a = 1
v = t + c
when t = 0 , v = 0 , thus c = 0
v = t
s = (1/2)t^2
so after 20 seconds,
s = (1/2)(400) = 200 m
v = 20 m/s (or 72 km/h )

second leg: at constant speed, no acceleration
v = 20 m/s
s = rate x time = 20(1800) = 36000 m

last leg:
a = ?
v = at + 20
when t = 30, v = 0, so
0 = 30a + 20
a = -1/15

s = (-1/15)t^2 + 20t

when t = 30
s = (-1/15)(900) + 30(20) = 540 m

so the total distance = 200 + 36000 + 540 m
= 36740 m
or
36.74 km

answer is reasonable, most of the trip is taken up by the 1/2 hr at a constant speed of 72 km/h which would be 36 km

Answer 2

first leg:
a = 1
v = t + c
when t = 0 , v = 0 , thus c = 0
v = t
s = (1/2)t^2
so after 20 seconds,
s = (1/2)(400) = 200 m
v = 20 m/s (or 72 km/h )

second leg: at constant speed, no acceleration
v = 20 m/s
s = rate x time = 20(1800) = 36000 m

last leg:
a = ?
v = at + 20
when t = 30, v = 0, so
0 = 30a + 20
a = -1/15

s = (-1/15)t^2 + 20t

when t = 30
s = (-1/15)(900) + 30(20) = 540 m

so the total distance = 200 + 36000 + 540 m
= 36740 m
or
36.74 km

answer is reasonable, most of the trip is taken up by the 1/2 hr at a constant speed of 72 km/h which would be 36 km

Answer 3

Good

Answer 4

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Answer 5

Kiss your left hand, say a weekday, put this in 15 other comment sections and he/she/ they will confess their feelings for you tomorrow.