Question

a particle is moving on a straight line, its velocity at time t is 8-2t ms/^2, what is the total distance covered from t=0 to t=6s

Answer 1

Given V = 8 - 2t;

velocity at t = 0s 
  $V_0$ = 8 - 2(0) = 8 $ms^{1}$

velocity at t = 6s
  $V_6$ = 8 - 2(6) = -4 $ms^{1}$

acceleration (a) = $\frac{-4-8}{6} = \frac{-12}{6} = -2 ms^{-2}; velocity = 0 [tex]ms^{-1}$ ⇒ t = 4s

distance travelled by the particle from t = 0 to t = 4
  S = ut + $\frac{1}{2}at^{2}$
⇒ S = 8(4) - $\frac{1}{2}(-2)4^{2} = 32 - 16 = 16 m$

distance traveled by the particle from t = 4 to t = 6
  S = ut + $\frac{1}{2}at^{2}$
⇒ S = 0(2) - $\frac{1}{2}(-2)2^{2} = 0 - 4  = -4 m = 4 m$ in opposite direction.

Total distance traveled by the particle = 16 m + 4 m = 20 m

Total displacement of the particle = 16 m - 4 m = 12 m

 

Answer 2

given,
t=0 sec
so, t=8-2t ms/2
t=8-2×0/2
= 0/2
=0 ans

step 2-
t=6sec
=8-2×6/2
=6×6/2
=36/2
=18 ans