Question

A Racing bike has uniform acceleration of 20 metre per second square what distance will it cover in 10 second after it starts.

Answer 1

To Find :-

The distance covered by the bike.

Given :-

• Acceleration of the Car = 20 m/s².

• Time taken = 10 s

We Know :-

Second Equation of Motion :-

$\bf{\underline{S = ut + \dfrac{1}{2}at^{2}}}$

Where ,

• S = Distance Traveled
• u = Initial velocity
• a = Acceleration produced
• t = Time Taken

Concept :-

According to the question ,

• The initial velocity of the bike will be Equal to zero , as it is starting from rest .ie, (u = 0).

• The Acceleration will be positive as the bike is accelerating itself.

Solution :-

Using the Second Equation of motion and by Substituting the values in it , we get :

$\implies \bf{S = ut + \dfrac{1}{2}at^{2}} \\ \\ \\ \implies \bf{S = 0 \times 10 + \dfrac{1}{2} \times 20 \times 10^{2}} \\ \\ \\ \implies \bf{S = \dfrac{1}{2} \times 20 \times 100} \\ \\ \\ \implies \bf{S = 10 \times 100} \\ \\ \\ \implies \bf{S = 1000 m} \\ \\ \\ \therefore \purple{\bf{S = 1000 m}}$

Hence, the distance covered by the bike is 1000 m.

Answer 2

Given :

• Initial velocity (u) = 0 m/s
• Time (t) = 10 seconds
• Acceleration (a) = 20 m/s²

To find :

• The distance (s)

In this question,

We will use Newton's eqation of motion

$\star \boxed{ \sf \purple{s = ut + \frac{1}{2} {at}^{2} }}$

Now,

According to the formula we will put the values,

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

$\sf{ = 0 \times 10 + \frac{1}{2} \times 20 \times {(10)}^{2} }$

$\sf{ = 0 + \frac{1}{ \cancel2} \times \cancel{20} {}^{ \: 10} \times 100}$

$\sf{ =1000 \:m }\orange\bigstar$

So,the distance is 1000 m...