A particle is moving with a constant acceleration of 4.0 m/s 2 . Its speed at t

Question

1. A particle is moving with a constant acceleration of 4.0 m/s2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? a. 8.0 m b. 8.0 m/s c. 16 m d. 16 m/s2 e. 12 m

1. A particle is moving with a constant acceleration of 4.0 m/s2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? a. 8.0 m b. 8.0 m/s c. 16 m d. 16 m/s2 e. 12 m

1. A particle is moving with a constant acceleration of 4.0 m/s2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? a. 8.0 m b. 8.0 m/s c. 16 m d. 16 m/s2 e. 12 m

A particle is moving with a constant acceleration of 4.0 m/s2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? a. 8.0 m b. 8.0 m/s c. 16 m d. 16 m/s2 e. 12 m A particle is moving with a constant acceleration of 4.0 m/s2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? A particle is moving with a constant acceleration of 4.0 m/s2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? 8.0 m 8.0 m 8.0 m/s 8.0 m/s 16 m 16 m 16 m/s2 16 m/s2 12 m 12 m a. 8.0 m b. 8.0 m/s c. 16 m d. 16 m/s2 e. 12 m

Explanation / Answer

The general equation of motion is :

x = ut + 0.5*at^2

where u = initial velocity

a = 4 m/s2

So, v = dx/dt = u + at

At , t = 1, v = 4m/s

So, 4 = u + 4*1

So, u = 0

So, the equation of motion,

x = 0 + 0.5*a*t^2 = 0.5*at^2

So, for area under position(x)- time(t) graph is given by the deficite integral of x and t from t = 1 to t=3

So, area = 1[ 0.5*a*t^3/3 ]3 = 0.5*a*(3^3 - 1^3)/3 = 0.5*4*(27-1)/3 = 17.3 m.s

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