Question

A snowmobile 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 1

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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

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 

Answer 2

$\sf \purple{ \clubs \: Given : }$

$\bf\color{magenta}\underline{Case \: 1 : }$

A snowmobile has an initial velocity of 4.7 m/s and it accelerates at the rate of 0.73 m/s² for 5.6 s .

$\bf\color{magenta}\underline{Case \: 2 : }$

It accelerates at the rate of -0.51 m/s².

$\sf \purple{ \clubs \: To \: find : }$

$\bf\color{magenta}\underline{Case \: 1 : }$

Final velocity of snowmobile .

$\bf\color{magenta}\underline{Case \: 2 : }$

How long it will take to reach a complete stop ?

$\sf \purple{ \clubs \: Solution : }$

$\bf\color{magenta}\underline{Case \: 1 : }$

Initial Velocity (u) = 4.7 m/s

Acceleration (a) = 0.73 m/s²

Time (t) = 5.6 s

Final Velocity (v) = ?

Formula used :

v = u + at

$\longmapsto\tt \: v \: = 4.7 + 0.73 \times 5.6 \\ \longmapsto\tt \: v \: = 4.7 +4.088 \: \: \: \: \: \: \: \: \: \: \\ \longmapsto\tt \: v \: =8.788 \: m {s}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{\boxed{ \sf \: Final \: Velocity \: (v) =8.788 \: m {s}^{ - 1} }}$

$\bf\color{magenta}\underline{Case \: 2 : }$

Initial Velocity (u) = 4.7 m/s

Acceleration (a) = -0.51 m/s²

Final Velocity (v) = 0 ( At stop )

Time (t) = ?

Formula used :

v = u + at

$\longmapsto\tt 0 = 4.7 + ( - 0.51)t \\ \longmapsto\tt4.7 + ( - 0.51)t = 0 \\\longmapsto \tt4.7 - 0.51t = 0 \: \: \: \: \: \: \\ \longmapsto\tt \cancel- 0.51t = \cancel- 4.7 \: \: \: \: \: \: \: \\\longmapsto \tt 0.51t = 4.7 \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\\longmapsto \tt \: t = \frac{4.7}{0.51} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\longmapsto \tt \: t \approx9.21 \: s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{\boxed{ \sf Time \: taken \: to \: reach \: complete \: stop (t) \approx9.21 \:s }}$