Question

A particle is moving with a uniform acceleration. its position time take second is given in metres by the relation x = 3+4t+5t^2.Calculate the magnitude of its uniform acceleration.​

Answer 1

Given displacement=> x= 3+4t+5(t^2)

At t=0 x= 3+4(0) +5(0)= 3

So initial displacement is 3 units

At t=1 x= 12-3 = 9

At t=2 x= 31-3 = 28

If initial velocity is zero

Using s= ut+1/2at^2

At t=1 s= (0)(1)+1/2(a)(1)^2

9= a/2

a=18 units

At t=2 s= (0)(2)×1/2(a)(2)^2

a= 18 units

Answer 2

Here is ur answer:

Acceleration, a at any time t is,

a = dv/dt = d/dt{ 4+10t}

a= d(4)/dt + d(10t)/dt

= 0+10d(t)/dt

= 10×1.t^1-1 =10

a = 10×1

=10m/s^2