Plz help!! Physics numerical problems!!?

Question

Can anyone solve these problems for me:

1. A point initially at rest moves along x-axis.It's acceleration varies with time as a=(6t+5) m/(s^2) . If it starts from the origin, find the distance covered by the particle in 2 second.

2. The height Y and distance x along the horizontal plane of a projectile on a certain planet are given by

Y=8t-5(t^2) and x=6t m. Find the velocity of the projectile??

3. the displacement x of a particle at time t along a straight line is given by

(x^2)=a(t^2)+ 2bt + c

where a,b,c are constants.Find the acceleration of the particle??

Answer 1

<< A point initially at rest moves along x-axis.It's acceleration varies with time as a=(6t+5) m/(s^2) . If it starts from the origin, find the distance covered by the particle in 2 second. >>

acceleration = change in velocity per unit time

a = dv/dt = 6t + 5

v = integral (6t + 5) dt

v = 3t^2 + 5t + C

where C = constant of integration and since point was initially at rest (given), C = 0 hence

v= 3t^2 + 5t

v = distance/time = ds/dt = 3t^2 + 5t

s = integral (3t^2 + 5t)dt

s = t^3 + 5t^2/2 + C

Since at t = 0, the point is at the origin (given), then C = 0 and

s =t^3 + 5t^2/2

and at t=2,

s = (2^3) + 5(2^2)/2 = 8 + 10

s = 18

<< The height Y and distance x along the horizontal plane of a projectile on a certain planet are given by

Y=8t-5(t^2) and x=6t m. Find the velocity of the projectile >>

From the given equations for Y and X, the following are evident for projectile analysis:

Vy = vertical component of the velocity = 8

Vx = vertical component of the velocity = 6

Therefore, the projectile velocity is

V^2 = 6^2 + 8^2

V^2 = 36 + 64 = 100

V = 10

and its direction is

tan A = Vy/Vx = 8/6

A = arc tan 8/6

A = 53.13 degrees



<<  the displacement x of a particle at time t along a straight line is given by (x^2)=a(t^2)+ 2bt + c where a,b,c are constants.Find the acceleration of the particle? >>

The acceleration of a particle is the second derivative of the displacement equation. Hence, if the original displacement equation is

x^2=a(t^2)+ 2bt + c

then the velocity (first derivative) is

2x(dx/dt) = 2a(t) + 2b

Solving for (dx/dt),

dx/dt = (at + b)/x

Taking the derivative of (dx/dt),

acceleration = d^2x/dt^2 = [x(a) - (at + b)(dx/dt)]/x^2

Simplifying the above since (dx/dt) = (at + b)/x

d^2x/dt^2 = [xa - ((at + b)(at + b)/x)]/x^2

d^2x/dt^2 = (xa - (at + b)^2/x)/x^2

d^2x/dt^2 = [(x^2)a - (at + b)^2]/x^3