Question

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 16.0 s. 2. Maintain a constant velocity for the next 1.20 min. 3. Apply a constant negative acceleration of ?9.34 m/s2 for 4.54 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

a = 2.65 m/s²;

t1 = 16 s;

t = 1.20 min = 72 sec;

a' = -9.34 m/s²

t3 = 4.54 s

(a)

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

V = u + at = 0 + a*t1

V = 2.65 x 16 = 42.4 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 2.65 x (16)²) + [42.4 x 72 + 0] + ( 42.4 x 4.54 + 1/2 x (-9.34) x (4.54)²]

s(total) = 339.2 + 3052.8 + 96.24 = 3488.24 meters

s(total) = 3.488 Km.

(b)

Speed for leg 1 will be

v1 = s1 / t1 = 339.2 / 16 = 21.2 m/s

Speed for leg 2 will be

v2 = s2 / t2 = 3052.8 / 72 = 42.4 m/s

Speed for leg 3 will be

v2 = s3 / t3 = 96.24 / 4.54 = 21.2 m/s

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

V = s(total) / t(total)

t(total) = t1 + t2 + t3 = 16 + 72 + 4.54 = 92.54 s

V = s(total) / t(total) = 3488.24 m / 92.54 = 37.7 m/s

Hence, V = 37.7 m/s