1. a surfer moving with an initial speed of 1.5 m/s catches a wave and accelerates at a constant
rate of 2.1 m/s2 for 2.2 s.
a) how far does the surfer travel in those 2.2 seconds?
b) what's the velocity of the surfer at the end of those 2.2 seconds?

a) Δx ≈ 8.4 m

b) v ≈ 6.1 m/s

Explanation:

Given:

v₀ = 1.5 m/s

a = 2.1 m/s²

t = 2.2 s

a) Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.5 m/s) (2.2 s) + ½ (2.1 m/s²) (2.2 s)²

Δx ≈ 8.4 m

b) Find: v

v = at + v₀

v = (2.1 m/s²) (2.2 s) + (1.5 m/s)

v ≈ 6.1 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s