mooncake9090
31.07.2020 •
Mathematics
A,B and C are three points lying in that order on a straight line. A body is projected from B towards A with speed 3m/s. The body experiences an acceleration of 1m/s² towards C. If BC=20m, find the time taken to reach C and the distance travelled by the body from the moment of projection until it reaches C.
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Ответ:
The total time taken to reach C = 10 s
The total distance traveled by the body = 29 m
Step-by-step explanation:
Whereby the points are vertically oriented, we have;
v = u + at
Where;
a = Acceleration of the body = -1 m/s²
u = Initial velocity = 3 m/s
v = Final velocity, = 0 m/s at maximum height
t = Time taken
0 = 3 + (-1)× t
t = 3 s
The maximum height S is given as follows;
S = u·t + 1/2·a·t² = 3×3 + 1/2×(-1)×3² = 4.5 m
Therefore, the total distance, , from maximum height to C = 45 + 20 = 24.5 m
The total time is given from the total distance as follows;
= u·t + 1/2·a·t²
Where;
u = Initial velocity from the maximum height = 0 m/s
a = Downward acceleration = Opposite sign to upward acceleration = 1 m/s²
24.5 = 0·t + 1/2·(1)·t²
49 = t²
t = √ 49 = 7 s
The total time taken to reach C = Initial time to reach maximum height + Time from maximum height = 3 s + 7 s = 10 s
The total time taken to reach C = 10 s
The total distance traveled by the body = Distance from B to maximum height + Distance from maximum height to C
The total distance traveled by the body = 4.5 m + 24.5 = 29 m.
The total distance traveled by the body = 29 m.
Ответ:
318 mm2
Step-by-step explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!