The speed of a particle after t seconds is given by v = 7 + 25t - 4t^2. Find the acceleration of the particle after 2 seconds?

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Answer 1 · 2018-02-18T17:58:10

We know,acceleration = #(dv)/(dt)# i.e rate of change in velocity

given, #v= 7+25t+4t^2#

So, #(dv)/(dt) =25 + 8t#

So, #|(dv)/(dt)|_2 = 25 +8*2 =41 ms^-2#

Answer 2 · 2018-02-19T14:12:30

I get #9 \ "m/s"^2#.

Acceleration is defined as the rate of change of velocity over time, in other words, we can write acceleration in terms of derivatives, such as

#a=(dv)/dt#, where #v# is the velocity of the object.

Here, we have #v=7+25t-4t^2#.

Using basic calculus rules, we get

#(dv)/dt=-8t+25#

Now, the question asks for the acceleration of the particle after #2# seconds, so we plug in #t=2#. Therefore,

#(dv)/dt=-8*2+25=9 \ "m/s"^2#

#<=>a=9 \ "m/s"^2#

So, the acceleration of the particle will be #9 \ "m/s"^2# after #2# seconds.