Question

At time t=0 a car has a velociry of 16m/s. it slows down with an acceleration given by -0.5t, in m/s² for t in seconds. it stops at t =? A.64s. B.32s. C.16s. D.8s. E.4s.​

Answer 1

Answer:

The answer is D.8s

Explanation:

We know,

$a=\frac{dv}{dt} \\=>adt=dv\\\\=>-0.5tdt=vdt ;[a=-0.5t]$.......(i)

For, t = 0, v = 16 ms⁻¹

      t = x, v = 0 ms⁻¹;(the car stops)

Now, Definite integral for equation (i),

$\int\limits^x_0 {-0.5t} \, dt=\int\limits^0_16{dx}$ ;[the lower limit will be 16 but I don't know why this is not working]

$=>-0.5(\frac{x^2}{2}-0)=-16$

∴$x=8s$

hope you have understood this...

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