Question

A bike is moving with a

velocity of 60m/s.The bike rider

applies break to produce a

uniform acceleration of -3 m/s2.

How far will the bike go before it

stops?​

Answer 1

Answer:

600 meters.

Explanation:

u = 60 m/s

v = 0 m/s

a = -3 m/s^2

s = ?

using the equation v^2 = u^2 + 2as

0 = 3600 - 6s

s = 600 meters.

So, he moves 600 meters before the bike stops.

Please mark it as brainliest.

Answer 2

Given :-

• Initial velocity, u = 60 m/s
• Retarded acceleration, a = -3 m/s²

To calculate :-

Distance travel before stopping ???

Solution :-

As the brakes are applied by the rider that means final velocity of the bike (v) would be = 0

Given, Retarded acceleration (a) = -3 m/s²

Here, negative sign indicates that the acceleration applied will bring the bike to the rest.

Now,

By using the laws of motion of equation

= v² - u² = 2as

Where,

s is total distance covered

v is the final velocity

a is the acceleration and

u is the initial velocity

After putting all the values;

==> 0 - 60² = 2 × (-3) × s

==> 0 - 60 × 60 = 2×(-3) × s

==> -3600 = -6s

After cancelling the negative sign;

==> s = 3600/6

==> s = 600m

Hence :-

The bike will travel 600m before it stops.