Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
