Question

An object, moving with constant velocity of 21 m/s travels for 4.5s. How far will it move during that time.

Answer 1

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Method 1:-

Here,

v = u = 21 m/s

t = 4.5 s,

We can also conclude that during the period, acceleration was = 0

[a = (v - u)/t = (v - v)/t = 0/t = 0]

We know,

s = ut + ½at²

or,

s = ut

⟹ s = (21 m/s)(4.5 s)

⟹ s = 94.5 m

∴ It will cover a distance of 94.5 m.

Method 2:-

We know,

s = vt

⟹ s = (21 m/s)(4.5 s)

⟹ s = 94.5 m.

Answer 2

GIVEN :–

• Initial velocity = Final velocity = 21 m/s [The velocity is constant]

• Time (t) = 4.5 seconds

• Acceleration = 0 m/s²

TO FIND :–

• The distance travelled by the object (s)

SOLUTION :–

Now we can easily find the distance covered by using the third equation of motion :—

• s = ut + ½at²

→ s = (21 × 4.5) + (½ × 0 × 4.5²)

→ s = 94.5 + 0

➠ s = 94.5 m

____________________________

ஃ The object travels 94.5 m in 4.5 seconds

ANSWER :–

• The distance travelled by the object = 94.5 m