Question

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.24 m/s2 for 10.0 s. 2. Maintain a constant velocity for the next 1.25 min. 3. Apply a constant negative acceleration of −9.86 m/s2 for 2.27 s. (a) What was the total displacement for the trip

Answer 1

Answer:

1817.448 m

Explanation:

Given:

Initial acceleration = 2.24 m/s²

Time for acceleration = 10 s

From  Newton's equation of motion

v = u + at

here v is the final velocity after the acceleration

v = 0 + 2.24 × 10 = 22.4 m/s

thus,

the displacement during acceleration

From  Newton's equation of motion

$s_1=ut+\frac{1}{2}at^2$

where,  

s is the distance

u is the initial speed = 0 m/s as starting from rest

a is the acceleration

t is the time

on substituting the respective values, we get

$s_1=0\times10+\frac{1}{2}\times2.24\times10^2$

or

s₁ = 112 m

for case 2

time = 1.25 minute = 1.25 × 60 = 75 seconds

displacement, s₂ = v × t = 22.4 × 75 = 1680 m

for the case 3

acceleration = - 9.86 m/s²

now,

the final velocity for the second case is the initial velocity for this case

thus,

$s_3=22.4\times2.27+\frac{1}{2}(-9.86)\times2.27^2$

or

s₃ = 50.848 - 25.40 = 25.448 m

hence,

the total displacement = s₁ + s₂ + s₃ = 112 m + 1680 m + 25.448 m

= 1817.448 m