Question

A car starting from rest moves with uniform acceleration of 0.1ms/Square For 4min Find the speed and distance traveled

Answer 1

Given that, a car starting from rest means the initial velocity of the car is 0 m/s with a uniform acceleration of 0.1 m/s² for 4 min.

We have to find the speed and distance travelled by the car.

From above data we have, u = 0 m/s, a = 0.1 m/s² and t = 4 min = 240 sec.

Using the First Equation Of Motion,

v = u + at

Substitute the known values,

→ v = 0 + (0.1)(240)

→ v = 0 + 24.0

→ v = 24

Therefore, the final velocity of the car is 24 m/s.

Using the Third Equation Of Motion,

v² - u² = 2as

Substitute the values,

→ (24)² - (0)² = 2(0.1)s

→ 576 = 0.2s

→ 2880 = s

Therefore, the distance travelled by the car is 2880 m.

In second case, we have to find the distance travelled by the car.

Distance = Speed × Time

From the above calculations we have distance is 2880 m and time 240 sec (given in question).

Let us assume that the speed of the car is 'x' m/s.

→ 2880 = x × 240

→ 2880/240 = x

→ 12 = x

Therefore, the speed of the car is 12 m/s.

Answer 2

$\huge\sf\red{Answer:}$

Given:

⇒ A car starting from rest moves with uniform acceleration of 0.1ms/Square for 4 min.

Find:

⇒ Find the speed and distance travelled.

Using formula:

2nd equation of motion:

★ ${\sf{\underline{\boxed{\green{\sf{ s=ut+ \dfrac{1}{2}at^2}}}}}}$

Calculations:

⇒ $\sf s=0+ \dfrac{1}{2} (0.1)(240)^2$

⇒ ${\sf{\underline{\boxed{\green{\sf{24 - Equation (1) }}}}}}$

⇒ $\sf s= \dfrac{1}{2}(0.1)(57600)$

⇒ $\sf s=0.1(28800)$

⇒ ${\sf{\underline{\boxed{\green{\sf{s=2880 - Equation (2) }}}}}}$

Therefore, 2880 m is the distance travelled.

Using formula:

★ ${\sf{\underline{\boxed{\green{\sf{Speed=\dfrac{Distance}{Time} }}}}}}$

Adding values form equation (2), equation (1) - we get:

⇒ $\sf \dfrac{2880}{240}$

⇒ $\sf \dfrac{288}{24}$

⇒ ${\sf{\underline{\boxed{\green{\sf{12 \: m/s}}}}}}$

Therefore, 12 m/s is the speed.