Question

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:​

Answer 1

`\bf \underline{ \underline{Given : }}`

• Initial velocity,u = 0 m/s

• Final velocity,v = 60 m/s

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

`\bf \underline{ \underline{</p><p>To \: be \: calculated : }}`

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

`\bf \underline{ \underline{Solution : }}`

We will first calculate the acceleration of the jeep.

CASE 1 :

By Using first equation of motion ,

`\sf \: v = u + at`

`\sf \star \: Substituting \: the \: values...`

`\sf\rightarrow \: 60 = 0 + a * 300`

`\sf \rightarrow \: 60 = 300a`

`\sf \rightarrow \: 300a = 60`

`\sf \rightarrow \: a = \cancel (300)/(6)`

`\sf \rightarrow \: a = 50 \: m {s}^( - 2)`

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 ,

`\sf {v}^(2) = {u}^(2) + 2as`

`\sf \star \: Substituting \: the \: values...`

`\sf \rightarrow \: {60}^(2) = {0}^(2) + 2 * 50 * s`

`\sf \rightarrow \: 3600 = 100s`

`\sf \rightarrow \: 100s = 3600`

`\sf \rightarrow \: s = \cancel(3600)/(100)`

`\sf \rightarrow \: s = 36 \: m`

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