Asnowmobile has an initial velocity of 4.7 m/s.
if it accelerates at the rate of 0.73 m/s2 for 5.6s, what is the final velocity? answer in units of m/s, if instead it accelerates at the rate of -0.51 m/s2, how long will it take to reach a complete stop? answer in units of s

Q1. The answer is 8.788 m/s

V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time

We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?

V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s

Q2. The answer is 9.22 s

V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time

We have:
V2 = 0   (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?

0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2   where t1=temperature of cold reservoir ( ° k) = 135            t2=temperature of hot reservoir ( ° k)   = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58%  note: this is the theoretical maximum efficiency of a perfect heat engine.   real-life engines do not achieve this value.