Question

A particle starts from rest and its acceleration at any time t is given by a = (3 - 0.75 t) m/s2
Find its maximum velocity in m/s)!​

Answer 1

The maximum velocity is 6 m/s

Step-by-step explanation:

Given

The acceleration of the particle

$a=(3-0.75t)$ m/s²

We know that rate of change of velocity is acceleration

Therefore,

$a=\frac{dv}{dt}$

$\implies \frac{dv}{dt}=3-0.75t$

$\implies dv=(3-0.75t)dt$

$\implies \int_0^v dv=\int_0^t (3-0.75t)dt$

$\implies v=(3t-0.75\frac{t^2}{2})\Bigr_0^t$

$\implies v=3t-\frac{0.75t^2}{2}$

$\implies v=3t-\frac{3t^2}{8}$

$\implies v=-\frac{3}{8}(t^2-8t)$

$\implies v=-\frac{3}{8}(t^2-8t+16-16)$

$\implies v=-\frac{3}{8}(t^2-2\times 4t+4^2)+\frac{3}{8}\times 16$

$\implies v=-\frac{3}{8}(t-2)^2+6$

The velocity will be maximum when $t=2$ seconds

Maximum velocity

$v_{max}=6$ m/s

Hope this answer is helpful.

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