Question

A bike accelerates uniformly from rest to a speed of $7.10{\text{ m }}{{\text{s}}^{ - 1}}$ over a distance of $35.4$m. Determine the acceleration of the bike.
A. $0.412{\text{ m }}{{\text{s}}^{ - 1}}$
B. $0.512{\text{ m }}{{\text{s}}^{ - 1}}$
C. $0.612{\text{ m }}{{\text{s}}^{ - 1}}$
D. $0.712{\text{ m }}{{\text{s}}^{ - 1}}$

Answer 1

As the bike is accelerating informing. It can be calculated from the equation of motions.
Formula used: ${v^2} - {u^2} = 2as$
Where u represent initial speed
v represent final speed
a represent acceleration
s is the distance covered.

Complete step by step answer:
Now, as the bike acceleration from rest. So initial speed of bike, $u = 0m/s$ and the final speed of bike, $7.10m/s.$ The distance covered by bike, $s = 35.4m$
Now, as we know that
${v^2} - {u^2} = 2as$
Where u is the initial speed, v is the final speed
A is the acceleration
And s is the distance covered.
Putting these values, we know
${v^2} - {u^2} = 2as$
$\Rightarrow {\left( {7.10} \right)^2} - {0^2} = 2 \times a \times 35.4$
$\Rightarrow {\left( {7.10} \right)^2} = 2 \times a \times 35.4$
$\Rightarrow 50.41 = 70.8a$
$\Rightarrow a = \dfrac{{50.41}}{{70.8}}$
$\Rightarrow a = 0.712\,\,m{s^{ - 2}}$
So, the acceleration of the bike is $0.712\,\,m{s^{ - 2}}$.

So, the correct answer is “Option D”.

Additional Information:
Time taken, $t = \dfrac{{7.10}}{a} = \dfrac{{7.10}}{{0.712}} = 9.97\sec .$
It can also be calculated.

Note:
One can also solve it by using, $s = ut + \dfrac{1}{2}a{t^2}$ and $v = u + at$.
$35.4 = 0\left( t \right) + \dfrac{1}{2}a{t^2}$ and $7.10 = 0 + at$
$35.4 = \dfrac{1}{2}a{t^2}.........\left( 1 \right)$ and $at = 7.10$
$\Rightarrow t = \dfrac{{7.10}}{a}........\left( 2 \right)$
Put (2) is (1), we get
$35.4 = \dfrac{1}{2}a \times {\left( {\dfrac{{7.10}}{a}} \right)^2}$
$35.4 = \dfrac{{a \times 7.10}}{{2 \times {a^2}}}$
$\Rightarrow 35.4 \times 2 \times a = 50.41$
$\Rightarrow a = \dfrac{{50.41}}{{70.8}}$
$\Rightarrow a = 0.712\,\,m{s^{ - 2}}.$