Question

a car accelerates from 6M/S to 16M/S in 10 sec.calculate
A)the acceleration
B)the distance covered by the car in that time

Answer 1

$\huge\star{\blue{\underline{\mathfrak{Answer:}}}}$

Given:

Initial velocity = 6m/s

Final velocity = 16m/s

Time taken = 10 sec

To Find:

• Acceleration

• Distance covered by the train in that time

Solution:

Acceleration:

Using 1st equation of motion -

$\huge\bold\red{\mathfrak{v\;=\;u\;+\;at}}$

putting the given values in above equation we get:

$16 = 6 + a \times 10 \\ 16 = 6 + 10a \\ 16 - 6 = 10a \\ 10 = 10a \\ a = 1$

hence , acceleration is 1m/s^2

$\boxed{\boxed{\boxed{\mathbb{a=1m/{s}^{2}}}}}$

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•Distance travelled :

using second equation of motion,we get:

$\huge\bold\red{\mathbb{s=ut+\frac{1}{2}a{t}^{2}}}$

$s = 6 \times 10 + \frac{1}{2} \times 1 \times {10}^{2} \\ s = 60 + \frac{1}{2} \times 1 \times 100 \\ s = 60 + 50 \\ s = 110m$

hence , the distance covered is 110m

$\boxed{\boxed{\boxed{\mathbb{s = 110m}}}}$

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Answer 2

Given ,

Initial velocity (u) = 6 m/s

Final velocity (v) = 16 m/s

Time (t) = 10 sec

We know that , the first equation of motion is given by

$\large \sf \fbox{v = u + at}$

Thus ,

$\Rightarrow \sf 16 = 6 + a \times 10 \\ \\ \Rightarrow \sf 10 = 10a \\ \\\Rightarrow \sf a = \frac{10}{10} \\ \\ \Rightarrow \sf a = 1 \: \: m/ {s}^{2}$

$\therefore \sf \bold{ \underline{The \: acceleration \: is \: 1 \: m/ {s}^{2} }}$

And the second equation of motion is given by

$\large \sf \fbox{S = ut + \frac{1}{2}a {t}^{2} }$

Thus ,

$\Rightarrow \sf S = 6 \times 10 + \frac{1}{2} \times 1 \times {(10)}^{2} \\ \\ \Rightarrow \sf S = 60 + 50 \\ \\ \Rightarrow \sf S = 110 \: \: m$

$\therefore \sf \bold{ \underline{The \: distance \: is \: 110 \: m}}$