Question

A record of travel along a straight path is as follows:

1. Start from rest with constant acceleration of 2.88 m/s² for 19.0 s.
2. Maintain a constant velocity for the next 1.10 min.
3. Apply a constant negative acceleration of ?8.05 m/s² for 6.80 s.

(a) What was the total displacement for the trip?
m

(b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?

leg 1	m/s
leg 2	m/s
leg 3	m/s
complete trip	m/s

Answer 1

Given that,

a = 2.88 m/s² ; t1 = 19 s ; t = 1.10 min = 66 sec ; a' = -8.05 m/s² t3 = 6.8 s

(a)The velocity at the end of the first interval will be given by

V = u + at = 0 + a t1 = 2.88 x 19 = 54.72 m/s

As per the point (2) this velocity will remain same for second interval and will be the intial velocity for the third interval.

Now, we know that, s = u t + 1/2 a t² where s is the displacement.

In our case, the total displacement will be the sum of displacements in the given three intervals, so

s(total) = s1 + s2 + s3

s(total) = [0 + 1/2 x 2.88 x (19)²) + [54.72 x 66 + 0] + ( 54.72 x 6.8 + 1/2 x (-8.05) x (6.8)²]

s(total) = 519.84 + 3611.52 + 185.98 = 4317.34 meters

Hence,s(total) = 4317.34 meters = 4.32 Km.

(b)Speed for leg 1 will be

v1 = s1 / t1 = 519.84 / 19 = 27.36 m/s

Speed for leg 2 will be

v2 = s2 / t2 = 3611.52 / 66 = 54.72 m/s

Speed for leg 3 will be

v2 = s3 / t3 = 3611.52 / 66 = 185.98 / 6.8 = 27.35 m/s

The average velocity of the complete trip will be given as follows:

V = s(total) / t(total)

t(total) = t1 + t2 + t3 = 19 + 66 + 6.8 = 91.8 s

V = s(total) / t(total) = 4317.34 meters / 91.8 = 47.03 m/s

Hence, V = 47.03 m/s