Question

pls help me out with this question guys....​

Answer 1

Answer:

1

Explanation:

v= 10 e^(-t/2)

a = dv/dt = 10 e^(-t/2) * (-1/2)

= -5e^(-t/2)

This means acceleration starts from -5 when time is 0 .

As time increases the value of e^(-t/2) will get smaller making -5e^(-t/2) less negative so B.

Answer 2

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$\large\mathcal\red{solution}$

now ..velocity is given by ..,

$v = 10e { }^{ \frac{ - t}{2} } \\ acceleration = \frac{dv}{dt} = 10 \times ( \frac{ - 1}{2} )e {}^{ \frac{ - t}{2} } \\ = > a = \frac{dv}{dt} = - 5e { }^{ \frac{ - t}{2} } \\ \\ now \: at \: t = 0 \\ a = - 5e {}^{0} = - 5 \: \: m.s {}^{ - 2}$

so at t=0 its acceleration was -5m/sec^2 ..i.e., retardation.....now it decreases exponentially with increase in time ....

so ... option (a) is correct.

$\underline{\large\mathcal\red{hope\: this \: helps \:you......}}$