Question

The velocity v(t) of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s2)t - (6.2 m/s3)t2. What is the average acceleration of the particle between t = 1.0 s and t = 2.0 s (up to two significant figures) ?

A) -13 m/s2

B) -14 m/s2
C) 13 m/s2
D) 15 m/s2
E) None of the above

Answer 1

Answer:

average acceleration = −14.5 m/s²

so correct option is B) -14 m/s²

Explanation:

given data

v(t) = (2.3 m/s) + (4.1 m/s2)t - (6.2 m/s3)t²

t1 = 1.0 s

t2 = 2.0 s

to find out

What is the average acceleration

solution

we know that average rate of change on the interval is

average acceleration = $\frac{Vf-Vi}{t2 - t1}$    ............1

so here put value of t for final f and initial i

we get Vf =  2.3 + (4.1)(2) - (6.2)(2)²

and Vi = 2.3 + (4.1)(1) - (6.2)(1)²

and t2 - t1 = 2 - 1 = 1 s

put value in equation 1

average acceleration = [2.3 + (4.1)(2) - (6.2)(2)² ] - [ 2.3 + (4.1)(1) - (6.2)(1)² ]

average acceleration = −14.5 m/s²

so correct option is B) -14 m/s²