Question

A 2650 kg truck traveling east at 39 km/h turn south and accelerates to 62 km/h I need help with part E, G, I & J

Answer 1

Ginve data:

The mass of truck is m=2650 kg.

The initial speed of truck is u=39 km/h (east).

The final speed of truck is v=62 km/h (south).

Part (e)

The velocity of truck in unit vector notation is,

`v=39(i)+62(-j)`

According to third law of Newton, the impulse and moment of a moving body is same, therefore we can write,

`\begin{gathered} p=mv \\ p=2650kg((39(km)/(h)*\frac{1\text{ m/s}}{(3.6km)/(h)})(i)+(62(km)/(h)*\frac{1\text{ m/s}}{(3.6km)/(h)})(-j)) \\ p=2650kg(((10.83m)/(s))(i)+((17.22m)/(s))(-j)) \\ p=(28708.3kgm)/(s)(i)+(45638.8kgm)/(s)(-j) \end{gathered}`

Thus, the x-component of impulse is 28708.2 kgm/s.

Part (g)

As calculated above the y-component of impulse is 45638.8 kgm/s.

Part (i)

The magnitude of impulse will be,

`\begin{gathered} p=\sqrt[]{28708.3^2+45638.8^2} \\ p=(53917.3kgm)/(s) \end{gathered}`

Thus, themagnitude of impulse is 53917.3 kgm/s.

Part (j)

The direction of impulse will be,

`\begin{gathered} \tan \theta=(45638.8)/(28708.3) \\ \theta=57.8\degree \end{gathered}`

The direction counterclock wise from x-axis will be,

`\begin{gathered} \theta=270\degree+57.8\degree \\ \theta=327.8\degree \end{gathered}`