Question

A body starts from rest and accelerates uniformly for 30seconds.if the acceleration is sms–² calculate the distincets covered by the body(hint÷s=ut+½at²

Answer 1

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From the Question,

• Initial Velocity,u = 0 m/s

• Time taken,t = 30s

• Acceleration,a = 5 m/s²

To find:

Distance travelled by the object

Using the Relation,

$\large{ \sf{s = ut + \frac{1}{2}at {}^{2} }}$

Since,the body is starts from rest

$\boxed{\boxed{\large{ \sf{s = \frac{1}{2}at {}^{2}}}}}$

Putting the values,we get:

$\implies \: \sf{s = \frac{1}{2} \times 5 \times 30 \times 30 } \\ \\ \implies \: \sf{s = 15 \times 5 \times 30} \\ \\ \implies \: \underline{ \boxed{ \sf{s = 2250 \: m}}}$

Thus,the object covers a distance of 2250m in 30 seconds with a velocity of 5m/s²

Answer 2

Correct Question :

A body starts from rest and accelerates uniformly for 30 seconds. If the acceleration is 5 m/s². Calculate the total Distance covered by the body.

AnswEr :

• Initial Velocity ( u ) = 0 m/sec
• Time ( t ) = 30 sec
• Acceleration ( a ) = 5 m/sec²
• Distance Travelled ( s ) = ?

• We will Use this Formula to calculate :

⇒ s = ( u × t ) + ( ½ × a × t² )

• Plugging the Values

⇒ s = ( 0 × 30 ) + ( ½ × 5 × (30)² )

⇒ s = 0 + ( ½ × 5 × 900 )

⇒ s = 0 + ( 5 × 450 )

⇒ s = 0 + 2250

⇒ s = 2250 m or, 2.25 Km

჻ Total Distance Travelled by the Body is 2250 m or 2.25 Km