Answer:
The given equation, x=10t+15t3 and the variables are (i) t = 0 to t = 2 s (ii) t = 2 s
We need to clarify all the variables.
Velocity of particle, v=dxdt
While calculating instantaneous velocity , we need to differentiate. Differentiating x w.r.t. time.
v=ddt(10t+15t3)=10+45t2
At t=0,V0=10+45(0)=10ms−1
At t= 2 s,
V2=10+45×(2)2=10+180=190ms−1
ΔV=V2−V1=190−10=180ms−1
Δt=2−0=2s
∴aav=ΔVΔt=1802=90ms−2
Now, differentiate, v with respect to time ti final the acceleration of the particle.
a=ddt(10+45t2)=90t
At t=2s,a=90×2=180ms−2
The instantaneous acceleration of a particle at 2 is 180ms−2.
Author:
sumodr1a
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0Answer:
you mentioned the value of y,ok
but i can't find x?
so,plz write complete question or attach it's photo
if you want the value of a then it is
a=t²+2t-5
a=1²+2×1-5
a=1+2-5
a=3-5
a= -2
ok bye .
Author:
milorksz
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