Question

2.
The position x of a particle varies with time t as
x = 6 + 12t – 2t where x is in metre and t in seconds.
The distance travelled by the particle in first five seconds
(a) 16 m
(b) 26 m
(c) 10 m
(d) 36 m​

Answer 1

Answer:

50 metre

Explanation:

putting limit from t=0 to t=5s

we get

x=50 metre

Answer 2

Answer:

Answer:

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

x ^ 0 = 6 + 12 × 0 - 2 × 0²

x ^ 0 = 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

x ^ 5 = 6 + 12 × 5 – 2 × 5²

x ^ 5 = 6 + 60 - 50

x ^ 5 = 16 m

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

∆ x = | x³ - x ^ 0 |

∆ 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'

= 18 + 8

= 26 m

Therefore, Total Distance travelled by Particle in 5 s is 26 m.