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Question: The position of a point in time 't' is given by \(x = a + bt - ct^{2}\), \(y = at + bt^{2}\). Its ac...

The position of a point in time 't' is given by x=a+bt−ct2x = a + bt - ct^{2}, y=at+bt2y = at + bt^{2}. Its acceleration at time 't' is

A

b−cb - c

B

(b+c)(b + c)

C

2b−2c2b - 2c

D

2b2+c22\sqrt{b^{2} + c^{2}}

Answer

2b2+c22\sqrt{b^{2} + c^{2}}

Explanation

Solution

Acceleration in x-direction = d2xdt2=−2c\frac{d^{2}x}{dt^{2}} = - 2c and acceleration in y-direction = d2ydt2=2b\frac{d^{2}y}{dt^{2}} = 2b

Resultant acceleration is = (−2c)2+(2b)2\sqrt{( - 2c)^{2} + (2b)^{2}} = 2b2+c22\sqrt{b^{2} + c^{2}}