Question

A car initially at rest starts moving with a constant acceleration of 0.8 ms-2 and travels a
distance of 40 m. Find its final velocity.
O m/s
O 7m/s
O 8m/s​

Answer 1

Explanation:

As, the car is initially at rest.

Initial velocity = 0m/s

Acceleration = 0.8m/s²

Distance = 40m

Now, Using the 2nd equation of motion:-

⇢ v² - u² = 2as

Here:-

• v = final velocity = ?

• u = initial velocity = 0m/a

• a = Acceleration = 0.8m/s²

• s = distance = 40m

⇢ v² = 2as + u²

⇢ v² = (2 × 0.8 × 4) + 0²

⇢ v² = 2 × 8 × 4

⇢ v² = 64

⇢ v = √64

⇢ v = 8

∴ Final velocity = 8m/s

Option (iii) is correct

Other equations of motion:-

1) v = u + at

2) v² - u² = 2as

3) s = ut + ½ at²

4) S = $\rm \dfrac{(u + v)t}{2}$

Answer 2

Given :-

☄ Initial velocity, u = 0m/s

☄ Acceleration, a = 0.8m/s²

☄ Distance travelled, s = 40m

To find :-

final velocity of the car, v

Solution :-

Using the equation of motion .i.e.,

➠ v² - u² = 2as

➠ v² - 0² = 2(0.8)(40)

➠ v² = 64m/s

➠ v = √64 m/s

➠ v = 8m/s

thus, Final velocity of the car is 8m/s.

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