## A record of travel along a straight path is as follows: (a) Start from rest with constant acceleration of 3.9 m/s 2 for 15.3 s; (b) Constant

Question

A record of travel along a straight path is as follows: (a) Start from rest with constant acceleration of 3.9 m/s 2 for 15.3 s; (b) Constant velocity of 59.67 m/s for the next 0.934 min; (c) Constant negative acceleration of −11.1 m/s 2 for 3.71 s. What was the total displacement

in progress 0
3 years 2021-08-06T12:01:53+00:00 2 Answers 9 views 0

a) 456.48m

b) 3343.91m

c) – 76.376m

Total displacement = 3724.01m

Explanation:

Using kinematic equation for displacement

◇x1 = Vo t + 1/2 at^2

Where ◇x1 = displacement

Vo = initial velocity

a = acceleration

t = time

◇x1 = 0(15.3) + 1/2 (3.9)(15.3)^2

◇x1 = 0 + 456.48

Displacement = 456.48m

b) ◇x2 = Vt1

V = velocity = 59.67m/s

◇x2 = 59.67 ×0.934minute×60seconds

◇x2 = 59.67 ×56.04 = 3343.91

c)

a= -11.1m/s

= (-11.1 m/s^2× 3.71seconds)

◇ x3 =1/2 (41.18m/s ×3.71)

◇x3 =- 76.376m

Total displacement = ◇x1 + ◇x2 + ◇x3

Total displacement = 456.48 + 3343.91 – 76.376

Total displacement = 3724.01m

Total displacement Is 1410.9945 meters

Explanation:

The total displacement is as follows.

The first part = triangle A

3.9 m/s² for 15.3 sec.

3.9 * 15.3 = 59.67 m/s

Second part = rectangle

15.67 m/s for (0.934*60)sec= 56.04 sec

Third part = triangle B

11.1m/s² for 3.71 sec

11.1 * 3.71 = 41.171 m/s

The area if these shapes gives the displacement.

For A = 1/2(15.3*59.67)

= 456.4755 m

For B = 15.67 * 56.04

= 878.1468 m

For C = 1/2(41.171*3.71)

= 76.372

Total displacement = 456.4755 + 878.1468 + 76.3722

= 1410.9945 meters