Question

The x and y coordinates of a particle at any time t are x
= 5t - 3t2 and y = 5t respectively, where x and y are in
meter and t in second. The speed of the particle at t =
1 second is
26 m/s
VT8 m/s
6 m/s
26 m/s​

Answer 1

Answer:

-4ms²

Explanation:

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Answer 2

SOLUTION :-

GIVEN :-

The x and y coordinates of a particle at any time t are

$\sf{x = 5t - 3 {t}^{2} \: \: and \: \: y = 5t } \: \: respectively$

where x and y are in meter and t in second.

TO CHOOSE THE CORRECT OPTION :-

The speed of the particle at t = 1 second is

• 26 m/s

• √18 m/s

• 6 m/s

• √26 m/s

EVALUATION :-

Here the x and y coordinates of a particle at any time t are given by

$\sf{x = 5t - 3 {t}^{2} \: } \: \: .....(1)$

$\sf{ y = 5t } \: \: \: ......(2)$

So the equation are given in parametric form

Differentiating both sides of Equation (1) with respect to t we get

$\displaystyle \sf{ \frac{dx}{dt} = 5 - 6t}$

Putting t = 1 we get

$\displaystyle \sf{ \frac{dx}{dt} \bigg |_{t = 1} = 5 - 6 = - 1}$

Again Differentiating both sides of Equation (2) with respect to t we get

$\displaystyle \sf{ \frac{dy}{dt} = 5 }$

Putting t = 1 we get

$\displaystyle \sf{ \frac{dy}{dt} \bigg |_{t = 1} = 5 }$

Hence the required speed of the particle

at t = 1 second is

$\displaystyle \sf{ = \sqrt{ {( - 1)}^{2} + {(5)}^{2} } }$

$\displaystyle \sf{ = \sqrt{1 + 25} }$

$\displaystyle \sf{ = \sqrt{26} }$

FINAL ANSWER :-

The speed of the particle at t = 1 second is

√26 m/s

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