Question

A bike starting from rest on a lake accelerates in a straight line at a constant rate of 3 metre per second square for 5 s. How far does the bike travel during this time?​

Answer 1

Given :-

★ Acceleration, a = 3 m/s²

★ Time, t = 5 sec

★ Initial velocity, u = 0 m/s

( As bike starts from rest)

To Find :-

★ Distance , s

Solution :-

To find distance__

At first, we need to find final velocity.

So, we have to use 1 st equation of motion.

We know,

v = u +at

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

Substituting the values, we get

⟹ v = 0 + 3 × 5

⟹v = 15 m/s

Now

to find distance, we have to use 3rd equation of motion.

We know,

$\sf v^2-u^2=2as$

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

s = Distance

Substituting the values, we get

⟹ ( 15)² - (0)² = 2 × 3 × s

⟹ s = 37.5 m

Hence,

the distance covered by bike is = 37.5 m

Answer 2

✬ Distance = 37.5 m ✬

Explanation:

Given:

• Initial velocity of bike 0 m/s.( started from rest )
• Acceleration produced by bike is 3 m/s².
• Time is 5 seconds.

To Find:

• Distance travelled by bike ?

Formula to be used:

• s = ut + 1/2at²

Solution: Here we have

• u = 0 m/s
• s = ?
• t = 5 seconds
• a = 3 m/s²

Put the values on formula

$\implies{\rm }$ s = ut + 1/2at²

$\implies{\rm }$ s = 0(5) + 1/2(3)(5)²

$\implies{\rm }$ s = 0 + 3/2 $\times$ 25

$\implies{\rm }$ s = 0 + 75/2

$\implies{\rm }$ s = 37.5

Hence, bike travelled 37.5 m during this time.