# A snowmobile has an initial velocity of +4.4 m/s.If it accelerates at the rate of +0.34 m/s2 for 4.2 s, what is the final velocity?

If instead it accelerates at the rate of −0.78 m/s2, how long will it take to reach a complete stop?Answer in units of s.

Relevance

Use v = v₀ + a*t

Given v₀ = 4.4 m/s

then

a) when t = 4.2 s and a =0.34 m/s²,

v = 4.4m/s + 0.34m/s² * 4.2s = 5.8 m/s

b) and when a = -0.78 m/s² and v = 0 ("reach a ... stop")

0 = 4.4m/s - 0.78m/s² * t

t = 5.6 s

Both answers to two significant digits, matching the data.

Hope this helps!

• We can use v = u + at

so, v = 4.4 + 0.34(4.2) => 5.8 m/s

Again, using v = u + at we have,

0 = 4.4 + (-0.78)t

so, t = 4.4/0.78 => 5.6 seconds

:)>

• Vf = Vi+a*t = 4.4+0.34*4.2 = 5.8 m/sec

ts = (0-Vi)/a = -4.4/-0.78 = 5.6 sec

all answers with just 2 significant digits

• v = at + v₀

v = .34(4.2) + 4.4 = 5.828m/s (rounds to 5.8m/s)

Deceleration, using same formula:

0 = (-0.78)t +4.4 ((assuming 4.4 initial velocity))

t = 5.64103s (but rounds to 5.6s)