A man runs at a speed of 4m s

Asked by admin @ 20/08/2021 in Math viewed by 318 People

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 attime t

Answered by admin @ 20/08/2021

Answer:

The man shall gain the bus at t=4.3 s.

Step-by-step explanation:

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,

$s = v\times t$

$s =4t$ ......(1)

Using equation of motion for bus,

$s = d+ut+\dfrac{1}{2}at^2$

Substitute the values in the formula,

$s =6+\dfrac{1}{2}\times 1.2t^2$ .....(2)

Solving equation (1) and (2),

$4t-6=\dfrac{1}{2}\times1.2t^2$

$0.6t^2-4t+6=0$

Solving by quadratic formula,

$t=\frac{-(-4)\pm\sqrt{(-4)^2-4(0.6)(6)}}{2(0.6)}$

$t=\frac{4\pm\sqrt{1.6}}{1.2}$

$t=\frac{4+\sqrt{1.6}}{1.2},\frac{4-\sqrt{1.6}}{1.2}$

$t=4.38,2.27$

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.

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