Question

A car accelerates from 6ms-1 to 18ms-1 in 12 seconds. Assuming the acceleration to be uniform calculate a) the acceleration and b) the distance covered in 12 seconds.

Answer 1

• Acceleration produce by the car= 1 m/s²
• distance covered by the car in 12 seconds = 144 m

Explanation:

$\tt \large \purple{Given}$

• A car accelerates from 6ms-¹ to 18ms-¹ .
• Time taken by the car = 12 seconds .
• Acceleration is uniform .

$\\ \tt \large \purple{To \: find}$

• Acceleration produce by the car .
• The distance covered in 12 seconds.

$\\ \tt \large \purple{Formula\: used}$

$\rm acceleration \: = \dfrac{change \: in \: velocity}{time \: taken} \\ \\ \boxed{ \rm s = ut +\dfrac{1}{2} \: a{t}^{2} }$

$\\ \tt \large \purple{Solution}$

A car accelerates from 6ms-¹ to 18ms-¹ .

here, initial velocity of the car = 6 m/s

final velocity of the car = 18 m/s

$\rm acceleration \: = \dfrac{change \: in \: velocity}{time \: taken}$

$\rm \: acceleration \: = \: \dfrac{18 - 6}{12} \\ \\ \rm \: acceleration \: = \: \dfrac{12}{12} \\ \\ \underline{ \boxed{\rm \: acceleration \: = \: 1 \: {ms}^{ - 2} }}$

Now, Acceleration produce by the car is 1 ms-² .

$\\ \rm \: distance \: coverd \: by\: the\: car \:in \:12 \:seconds :-\\ \\ \rm \: s = ut +\dfrac{1}{2} \: a{t}^{2}\:$

$\\ \\ :\implies \rm \: s = ut +\dfrac{1}{2} \: a{t}^{2} \:$

$:\implies \rm \: here, \: \rm v \: = final \: velocity$

$:\implies \rm \: u \: = initial \: velocity$

$:\implies \rm a = acceleration$

$:\implies \rm t \: = time$

$:\implies \rm s = distance$

$:\implies \rm s = 6 \times 12 + \dfrac{1}{2} \times 1 \times {12}^{2}$

$:\implies\rm s \: = 72 + \dfrac{1}{2} \times 144$

$\\ \\ :\implies \rm s = 72 + 72 \\ \\ \underline{ \boxed{\rm s = 144 \: m}}$

Hence, distance covered by the car in 12 seconds is 144 m .

$\\ \\ \bf \underline{equation \: of \: motion \:} - \\ \red\bigstar \rm \: v \: =u + at \\ \\ \red\bigstar \rm \: s \: = ut + \frac{1}{2} \: a {t}^{2} \\ \\\red\bigstar \rm \: {v}^{2} = {u}^{2} + 2as$