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?
Answer in units of m/s.
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.
- NCSLv 71 month agoFavourite answer
Use v = v₀ + a*t
Given v₀ = 4.4 m/s
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!
- Wayne DeguManLv 71 month ago
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
- oubaasLv 71 month ago
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
- JimLv 71 month ago
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)