contestada

2a. A
becomes
jeep starts from it the
starts from it the state of rest. If its
velocity 60m/s and its take 5 minute: what is the accelera
ution of the jeep
the distanced travelled by
the jeep:​

Respuesta :

Itzisabella

[tex] \bf \underline{ \underline{Given : }}[/tex]

  • Initial velocity,u = 0 m/s

  • Final velocity,v = 60 m/s

  • Time taken,t = 5 min = 5 × 60 sec = 300 seconds

[tex] \bf \underline{ \underline{

To \: be \: calculated : }}[/tex]

Calculate the acceleration ( a ) and distance (s ) covered by the jeep.

[tex] \bf \underline{ \underline{Solution : }}[/tex]

We will first calculate the acceleration of the jeep.

CASE 1 :

By Using first equation of motion ,

[tex] \sf \: v = u + at[/tex]

[tex] \sf \star \: Substituting \: the \: values...[/tex]

[tex] \sf\rightarrow \: 60 = 0 + a \times 300[/tex]

[tex] \sf \rightarrow \: 60 = 300a[/tex]

[tex]\sf \rightarrow \: 300a = 60[/tex]

[tex]\sf \rightarrow \: a = \cancel \dfrac{300}{6} [/tex]

[tex]\sf \rightarrow \: a = 50 \: m {s}^{ - 2} [/tex]

Thus,the acceleration of the jeep is 50 m/s².

Now, Let us calculate the distance travelled by the jeep.

CASE 2 :

By Using third equation of motion ,

[tex] \sf {v}^{2} = {u}^{2} + 2as[/tex]

[tex] \sf \star \: Substituting \: the \: values...[/tex]

[tex]\sf \rightarrow \: {60}^{2} = {0}^{2} + 2 \times 50 \times s[/tex]

[tex]\sf \rightarrow \: 3600 = 100s[/tex]

[tex] \sf \rightarrow \: 100s = 3600[/tex]

[tex]\sf \rightarrow \: s = \cancel\dfrac{3600}{100} [/tex]

[tex]\sf \rightarrow \: s = 36 \: m[/tex]

Thus,the distance covered by the jeep is 36 m .

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