Question

a body starting with a velocity of 10m/s is accelerated uniformly from 60s . if the velocity acquired in this interval is 250m/s . find the acceleration and also the distance traveled ​

Answer 1

Answer:

$Acceleration \: = 4 \: ms {}^{ - 2} \\ Distance \: travelled \: by \: body = 7800m \:$

Explanation:

$Given:</p><p> \\ Initial \: velocity \: (u)= 10 \: ms {}^{ - 1} , \\ Final \: velocity \: (v)= 250 \: ms {}^{ - 1} , \\ Time \: (t) = 60 \: s \\ \\ Acceleration \: (a) = \frac{v - u}{t} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{250 - 10}{60} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{240}{60} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 4 \: ms {}^{ - 2} \\ \\ Distance \: travelled \: \: \: S=ut + \frac{1}{2} at {}^{2} \\ = 10 \times 60 + \frac{1}{2} \times 4 \times 3600 \\ = 600 + 2 \times 3600 \\ = 600 + 7200 \\ = > 7800 \: m$

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Answer 2

Answer:

Acceleration=4ms

−2

Distancetravelledbybody=7800m

Explanation:

\begin{gathered}Given: < /p > < p > \\ Initial \: velocity \: (u)= 10 \: ms {}^{ - 1} , \\ Final \: velocity \: (v)= 250 \: ms {}^{ - 1} , \\ Time \: (t) = 60 \: s \\ \\ Acceleration \: (a) = \frac{v - u}{t} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{250 - 10}{60} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{240}{60} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 4 \: ms {}^{ - 2} \\ \\ Distance \: travelled \: \: \: S=ut + \frac{1}{2} at {}^{2} \\ = 10 \times 60 + \frac{1}{2} \times 4 \times 3600 \\ = 600 + 2 \times 3600 \\ = 600 + 7200 \\ = > 7800 \: m\end{gathered}

Given:</p><p>

Initialvelocity(u)=10ms

−1

,

Finalvelocity(v)=250ms

−1

,

Time(t)=60s

Acceleration(a)=

t

v−u

=

60

250−10

=

60

240

=>4ms

−2

DistancetravelledS=ut+

2

1

at

2

=10×60+

2

1

×4×3600

=600+2×3600

=600+7200

=>7800m