A man runs at a speed of 4m/s to overtake a standing bus when he is 6 m bend the door the bus moves forward and continues with constant acceleration of 1.2 m/s.s the man shall gain the bus at times t
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Answer:
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Answer:
Given : A man runs at a speed of 4 m/s to overtake a standing bus when he is 6 m bend the door the bus moves forward and continues with constant acceleration of 1.2 m/s.
To find : The man shall gain the door at time t equal to ?
Solution :
Speed of man v= 4 m/s
Distance d = 6 m
Acceleration a=1.2 m/s²
Initial velocity u=0 m/s
Using the distance formula for man,
s = v x t
s = 4t ....(1)
Using equation of motion for bus,
s = d + ut + 1/2at^2
Substitute the values in the formula,
s = 6 + 1/2 x 1.2t^2
Solving equation (1) and (2),
4t - 6 = 1/2 x 1.2t^2
0.6t^2 - 4t + 6 = 0
Solving by quadratic formula,
t = -(-4) +- root (-4)^2-4(0.6)(6)
t = 4 +- root 1.6/1.2
t = 4 + root 1.6/1.2; 4 - root 1.6/1.2
t = 4.38; 2.27
At 2.27 s the velocity of the bus is greater than that of the man.
So man cannot catch the bus.
But after 4.3 s the relative velocity of the bus and man is zero so catching of the bus is possible.
Therefore, The man shall gain the bus at t=4.3 s.
Step-by-step explanation:
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