Question

the distance traveled by a particle moving along a straight line is given by x=(4t+5t^2+6t^3)m.
1)find the initial velocity of the particle.
2) the velocity at the end of 4 sec.
3) the acceleration of the particle at the end of 5 sec.

Original answers please!!!!!

Answer 1

Answer:

$\huge{\boxed{\boxed{Hello}}}$

first calculate the distance travelled by the particle.the particle second is 4. the acceleration of particle at the end of 5 second 16 metre answer

Answer 2

Answer:

100% correct answer

Explanation:

given: x= 4t + 5t^2 + 6t^3

we need to find v.

By using differential calculus,

v=dx/dt

v=d(4t + 5t^2 + 6t^3)/dt

* using formula, dn^x/dt = xn^x-1

v=d4t/dt + d5t^2/dt + d6t^3/dt

v=1×4t^1-1 + 2×5t^2-1 + 3×6t^3-1

v=4 + 10t + 18t^2

a) we need initial velocity, it means velocity at t=0

put t=0 in v=4+10t+18t^2

v= 4m/s

b) at t=4

put again the value and you will get v=332m/s

c) at t=5

v=504m/s

It took alot of time to type, plz like and vote 5