Kinematics

Question

A particle moves in a straight line so that , t seconds after passing through O, its velocity, v m/s , is given by v=t^2-8t+7. The particle comes to an instantaneous rest, firstly at A then at B. Find 

 

(a) an expression for the displacement , x m, of the particle from O at the time t,

(b) the distance AB

(c)the total distance travelled in the first 9 seconds after passing through O.

 

Given that C is the point at which the particle has zero acceleration, determine, with full working where C is nearer to O or to B.

Answer 1

a)

v(t) = t^2 - 8t + 7

x(t) = Integral of v(t)

x(t) = (1/3)t^3 - 4t^2 + 7t + constant

 

b)

 A & B are points in which the particle is at rest (v(t) = 0).

 

t^2 - 8t + 7 = 0

(t-7)(t-1) = 0

t=7

t=1

 

Therefore, the distance AB is x(7) - x(1) = (343/3 - 196 + 49) - (1/3 - 4 + 7) = 114 - 147 - 3 = -36

The distance of AB would be the absolute of the displacement which is 36 

 

c) You must calculate this in three parts since the particle changes direction twice.

 

x(1) - x(0) = 1/3 - 4 + 7 = 10/3

 

x(7) - x(1) = -36

 

x(9) - x(7) = (243 - 324 + 63) - (343/3 - 196 + 49) = 44/3

 

10/3 + 44/3 + 36 = 54