a=t2+2t-5,t=1 x=? physics numerical​

Answers 2

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.

Answer:

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 .

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