   Chapter 4.4, Problem 58E

Chapter
Section
Textbook Problem

# The acceleration function (in m/s2) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time t and (b) the distance traveled during the given time interval. a ( t ) = 2 t + 3 ,    v ( 0 ) = − 4 ,    0 ≤ t ≤ 3

To determine

To find:

The velocity at time t.

Explanation

1) Concept:

v't=a(t)

Therefore,vt=a(t)dt.

2) Given:

at=2t+3,  v0=-4,  0t3

3) Calculation:

Consider

at=2t+3

So

v't=at=2t+3

Integrating,

vt=(2t+3)dt

=2tdt+3dt

=2t22+3t+C

Therefore, vt=t2+3t+C where C is an arbitrary constant.

To solve for C Now, at t=0, v0=-4

Therefore,

02+30+C=-4

C=-4

Substitute in v(t).

Therefore,

vt=t2+3t-4 m/s

Conclusion:

Therefore, the velocity at time t is t2+3t-4 m/s

b)

To find:

The distance travelled during the given time interval.

Solution:

896 m

1) Concept:

Total distance travelled:t2t1v(t)dt ; t1tt2

2) Calculation:

From the part (a),

The velocity of the particle at time t is given by

vt=t2+3t-4

Thus, the distanced travelled is

03vtdt=03t2+3t-4dt

Since the velocity is negative on the given interval [0, 1] and positive on the interval [1, 3];

On the interval [0, 1], t2+3t-4=-t2+3t-4

On the interval [1, 3], t2+3t-4=+t2+3t-4

Therefore,

03t2+3t-4

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