Question

The position of particle varies with time as x= 4t^2 -8t + 2. Find the distance traversed by particle from t=0s to t=2s.

Answer 1

Given :

▪ Time dependent position equation of particle has been provided.

✴ x = 4t² - 8t + 2

To Find :

▪ Distance covered by body in first two seconds.

SoluTion :

Given : x = 4t² - 8t + 2

At,

t = 0s : x(0) = 4(0)² - 8(0) + 2 = 2m

t = 1s : x(1) = 4(1)² - 8(1) + 2 = -2m

t = 2s : x(2) = 4(2)² - 8(2) + 2 = 2m

↗ The distance covered by the particle in 1st second is

D¹ = |x(0) - x(1)| = |2 - (-2)| = 4m

↗ The distance covered by the particle in 2nd second is

D² = |x(1) - x(2)| = |(-2) - 2| = 4m

✴ Total distance covered by the particle in first 2 seconds is

D = D¹ + D²

D = 4 + 4

D = 8m

Answer 2

Answer:

8m

Explanation:

Here,

x = 4t^2 - 8t + 2

At,

t = 0s : x(0) = 4(0)^2 - 8(0) + 2 = 2m

t = 1s : x(1) = 4(1)^2 - 8(1) + 2 = -2m

t = 2s : x(2) = 4(2)^2 - 8(2) + 2 = 2m

=> The distance covered by the particle in 1st second is

D1 = |x(0) - x(1)| = |2 - (-2)| = 4m

=> The distance covered by the particle in 2nd second is

D2 = |x(1) - x(2)| = |(-2) - 2| = 4m

◆ Total distance covered by the particle in first 2 seconds is

D = D1 + D2

D = 4 + 4

D = 8m Ans...