Physics, asked by manandarak3766, 4 months ago

A body starting with a velocity of 10 metre per second is accelerated uniformly for 60sec. If the velocity acquired in this interval is 250 metre per second. Find the acceleration and also distance travelled

Answers

Answered by Cynefin

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

The above question is of Kinematics. Let's know the three equations of motion first of all,

v = u + at
s = ut + 1/2 at²
v² = u² + 2as

Here u is the initial velocity, v is the final velocity, a is the accleration, s is the distance travelled, t is the time taken.

GiveN:

u = 10 m/s
v = 250 m/s
t = 60 s

To FinD:

Accleration of the body?
Distance travelled by the body?

Step-by-step Explanation:

By using 1st equation of motion,

⇒ v = u + at

⇒ 250 = 10 + a(60)

⇒ 240 = 60a

⇒ a = 4 m/s²

Accleration of the body is 4 m/s² (Answer)

Now finding the distance travelled.

By using 2nd equation of motion,

⇒ s = ut + 1/2 at²

⇒ s = 10(60) + 1/2 (4)(60)²

⇒ s = 600 + 7200 m

⇒ s = 7800 m

Distance travelled by the body is 7800 m. (Answer)

Answered by Mister360

137

Answer:

$\huge \bf \: given$

A body starting with a velocity of 10 metre per second is accelerated uniformly for 60sec. If the velocity acquired in this interval is 250 metre per second. Find the acceleration and also distance travelled

$\huge \bf \: to \: find$

Acceleration and distance travelled by body

$\huge \bf \: solution$

According to first equation of motion

$\sf \: v = u + at$

$\sf \: 250 = 10 \: a(60)$

$\sf \: 240 = 60a$

$\sf \: a \: = 4m \: per \: {second}^{2}$

Distance travelled

According to second equation of motion

$\sf \: s= ut +\frac{1}{2} {a}^{2}$

${:}\longrightarrow$ $\sf \: s = 10 (60)+ \frac {1}{2} (4) (60²)$

${:}\longrightarrow$ $\sf \: s = 600 + 7200$

${:}\longrightarrow$ $\sf \: s = 7800$

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