Question

A bike at rest attains a velocity of 72km/h over a distance of 40m.Find the acceleration of the bike​

Answer 1

Solution:-

From the question,

Final velocity,v:72km/h

Distance:40m

→It is known that,the bike is at rest so the initial velocity is 0m/s. =>u=0m/s

Converting km/h into m/s:

72 km/h

=72×5/18

=4×5

=20m/s

Using the formula,

•v^2-u^2=2as

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

and s is the distance travelled.

Substituting the respective values,

(20)^2 - (0)^2=2(40)a

=>400=80a

=>8a=40

=>a=5

•Acceleration of the object is 5m/s^2

Answer 2

Answer:

5 m/s²

Step-by-step explanation:

Initial velocity of bike(u) = 0 m/s

Final velocity of bike (v) = 72 km/h

Converting km/h into m/s -

$\sf \implies 72km/h \\ \\ \sf \implies 72 \times \frac{5}{18} \\ \\ \sf \implies 72km/h = 20 \: m/s$

Distance travelled (s) = 40 m

To find: Acceleration of the bike.

At first, we need to find the time taken by bike to complete the given distance.

We will use the third equation of motion to find it.

v² - u² = 2as

⇒ (20)² - 0 = 2 * a * 40

⇒ 400 = 80 a

⇒ a = $\dfrac{400}{80}$

⇒ a = 5 m/s²

Hence, the answer is 5 m/s².