Question

The position of the point in time $'t'$is given by $x = a + bt - c{t^2},y = at + b{t^2}$. It’s acceleration at time $'t'$ is
\eqalign{ & 1.b - c \cr & 2.b + c \cr & 3.2b - 2c \cr & 4.2\sqrt {{b^2} + {c^2}} \cr}

Answer 1

There are two given equations at a given time $'t'$. We know that differentiating w.r.t. time will give velocity. Therefore, to find the acceleration, we need to differentiate twice. After the first differentiation, we get the velocity. Then, we differentiate for the second time, we arrive at the acceleration. The velocity and acceleration have two different formulas, we will differentiate once, but need not find the velocity. We will directly find out the acceleration.
Formulas used:
Velocity, $v = \sqrt {{{\left( {\dfrac{{dx}}{{dt}}} \right)}^2} + {{\left( {\dfrac{{dy}}{{dt}}} \right)}^2}}$
Acceleration, $a = \sqrt {{{\left( {\dfrac{{{d^2}x}}{{d{t^2}}}} \right)}^2} + {{\left( {\dfrac{{{d^2}y}}{{d{t^2}}}} \right)}^2}}$
$\dfrac{d}{{dx}}(const) = 0$
$\dfrac{d}{{dx}}({x^n}) = n{x^{(n - 1)}}$

Complete step-by-step solution:
Let us find out the first derivative of the equation, $x = a + bt - c{t^2}$, which will give the velocity,
Differentiating with respect to $t$,
$\dfrac{{dx}}{{dt}} = b - 2ct$
Now by differentiating again, we get the acceleration.
$\dfrac{{{d^2}x}}{{d{t^2}}} = - 2c$
Now consider the other equation, which is $y = at + b{t^2}$
Differentiating, with respect to $t$,
$\dfrac{{dy}}{{dt}} = a + 2bt$
This is the velocity, now we differentiate again to get the acceleration,
$\dfrac{{{d^2}y}}{{d{t^2}}} = 2b$
Now, we need to substitute the above values for the equation,
$a = \sqrt {{{\left( {\dfrac{{{d^2}x}}{{d{t^2}}}} \right)}^2} + {{\left( {\dfrac{{{d^2}y}}{{d{t^2}}}} \right)}^2}}$
\eqalign{ & \Rightarrow a = \sqrt {{{\left( { - 2c} \right)}^2} + {{\left( {2b} \right)}^2}} \cr & \Rightarrow a = \sqrt {4{c^2} + 4{b^2}} \cr & \Rightarrow a = 2\sqrt {{b^2} + {c^2}} \cr}
Therefore, the final answer, that is the acceleration at time $t$ is,
$a = 2\sqrt {{b^2} + {c^2}}$
Hence, option (4) is the correct answer.

Note: Pay attention to the equations given. There are two different equations and we need to differentiate each twice. So, we will be differentiating four times in total. Memorize the formulas for differentiation. Substitute proper formulas and make necessary simplifications. Also, note that there is a square root involved with the velocity and acceleration equations, ensuring to solve them steadily.