Question

A car accelerates uniformly from 18 km 36 km/h in 2 seconds. Calculate a. the acceleration b. the distance covered by the car in that time.

Answer 1

Answer:

Explanation:

Given that

Initial Velocity = 18 km/h = 18 * 5/18 = 5 m/s

Final velocity = 36 km/h = 36 * 5/18 = 10 m/s

time, t = 2 s

a) acceleration, a = (v-u)t = 10 - 5 / 2 = 5/2 = 2.5 m/s^2

b) distance, s is given by

$\frac{v^{2} -u^{2} }{2a} \\= \frac{10^{2}-5^{2} }{2*2.5} \\= 100-25 / 5\\= 75/5\\= 15 m$

Answer 2

GIVEN:

• Initial velocity (u) - 18 Km/h
• Final velocity (v) = 36 Km/h
• Time taken (t) = 2 s

TO FIND:

• Acceleration (a)
• Distance (s)

SOLUTION:

From the first equation of motion :

• v = u + at

Where we have :

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time taken

Here :

⇒ v = 36 Km/h × 5/18

• 10 m/s

⇒ u = 18 Km/h × 5/18

• 5 m/s

⇒ t = 2 sec

Substituting we get :

• 10 = 5 + a(2)
• 10 = 5 + 2a
• 10 - 5 = 2a
• 5 = 2a
• a = 5/2
• a = 2.5 m/s²

From the third equation of motion :

• v² - u² = 2as

Where :

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

Here :

⇒ v = 36 Km/h × 5/18

• 10 m/s

⇒ u = 18 Km/h × 5/18

• 5 m/s

⇒ t = 2 sec

⇒ a = 2.5 m/s²

Substituting we get :

• 10² - 5² = 2(2.5)s
• 100 - 25 = 5s
• 75 = 5s
• s = 75/5
• s = 15 m

ANSWER:

• Acceleration (a) is 2.5 m/s².
• Distance covered is 15 m.

VERIFICATION:

• Using third law of motion.

⇒ v² - u² = 2as

Substituting the values,

• 10² - 5² = 2(2.5)(15)
• 100 - 25 = 5(15)
• 75 = 75

∴ L.H.S = R.H.S

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