Question

the velocity of a particle is given by v = t-4 where v is in meter / second, t in seconds. calculate distance covered by particle in 6 second

Answer 1

given: x=t^3-3t^2-10 --equation(1)

differentiating equation(1),we get v=3t^2-6t --equation(2)

 

at time t=0,from equation (1) the particle is at x=t^3-3t^2-10=0-0-10= -10. Now lets find out where the velocity is 0 i.e. its direction of motion is reversed(as it is given rectilinear motion). Setting equation(2) to 0, 3t^2-6t=0 =>3t(t-2)=0 or t=2 seconds. Therefore we have to calculate separately. At t=2, its position is x=t^3-3t^2-10= 2^3-3*2^2-10= 8-12-10= -14 . So from t=0 to t=2, distance travelled is x1=-10-(-14)=4 m. Now its position at t=4 ,x=x=t^3-3t^2-10=4^3-3*4^2-10=64-48-10=6, so from t=2 to t=4, it travelled x2=6-(-14)=6+14=20m.Therefore total distance travelled by the particle in 4 seconds starting from t=0 is x1+x2=4+20= 24m.

So ans is 24m.