Question

The position x of particle varies with time t as x = 6 + 12t – 2t²
where x is in metre and t in second. What
is the distance travelled by the particle in first 5 seconds?
plzz help me friends...

Answer 1

Dear Diya,

◆ Answer -

Distance = 26 m

◆ Explaination -

Position of the particle is given by -

x = 6 + 12t – 2t²

Differentiating w.r.t. t,

dx/dt = 12 - 2 × 2t

v = 12 - 4t

When particle stops travelling in forward direction, v = 0 -

0 = 12 - 4t

t = 12 / 4

t = 3 s

Position of the particle at t = 0 s,

x = 6 + 12t – 2t²

x0 = 6 + 12 × 0 - 2 × 0²

x0 = 6 m

Position of the particle at t = 3 s,

x = 6 + 12t' – 2t'²

x3 = 6 + 12 × 3 – 2 × 3²

x3 = 6 + 36 - 2 × 9

x3 = 24 m

Position of the particle at t = 5 s,

x = 6 + 12t' – 2t'²

x5 = 6 + 12 × 5 – 2 × 5²

x5 = 6 + 60 - 50

x5 = 16 m

Distance travelled by particle from t = 0 s to t = 3 s,

∆x = |x3 - x0|

∆x = |24 - 6|

∆x = 18 m

Distance travelled by particle from t = 3 s to t = 5 s,

∆x' = |x5 - x3|

∆x' = |16 - 24|

∆x' = 8 m

Therefore, total distance travelled by particle in 5 s will be -

Distance = ∆x + ∆x'

Distance = 18 + 8

Distance = 26 m

Therefore, total distance travelled by particle in 5 s is 26 m.

Thanks..