Question

If a body has a velocity 50 m s-1 and the uniform acceleration is 5 m s-2, find the time it takes to travel a distance of 240 m.
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Answer 1

Answer :-

Given :-

• Initial velocity = 50 m/s
• Acceleration = 5 m/s²
• Distance = 240 m

To Find :-

• Time

Solution :-

We know that,

→ s = ut + ½ at²

where

• s is distance
• u is initial velocity
• t is time
• a is acceleration

Substituting the values in formula :-

→ 240 = 50 × t + ½ × 5 × t²

→ 240 = 50t + ½ × 5t²

→ 240 = ( 100t + 5t² ) / 2

→ 240 × 2 = 100t + 5t²

→ 480 = 100t + 5t²

→ 480 = 5 ( 20t + t² )

→ 20t + t² = 480 / 5

→ 20t + t² = 96

→ t² + 20t - 96 = 0

→ t² + 24t - 4t - 96 = 0

→ t ( t + 24 ) - 4 ( t + 24 ) = 0

→ ( t - 4 ) ( t + 24 ) = 9

→ t = 4 , - 24

( As negative time is not possible, time can't be - 24 seconds. )

Time = 4 seconds

Answer 2

$\begin{gathered}\frak{Given}\begin{cases}\sf{ Initial~Velocity(u)=50m/s} \\\sf{Acceleration(a)=5m/s²}\\\sf{Distance(s)=240m}\end{cases}\end{gathered}$

To Find

• Time

We know that

• $\boxed{ \bf{s=ut+\dfrac{1}{2}at²}}$

Putting the value in Formula, we get:

$\sf→240=50t+\dfrac{5t²}{2}$

$\sf→240=\dfrac{100t + 5t²}{2}$

$\sf→ 5t² + 100t= 2 \times 240$

$\sf→ 5t² + 100t= 480$

$\sf→ 5t² + 100t - 480=0$

On Dividing both sides by 5, we get:

$\sf→ \dfrac{5t² + 100t - 480}{5}= \dfrac{0}{5}$

$\sf→ t² + 20t - 96= 0$

By Splitting middle term we get

$\sf→ t² + 24t - 4t - 96= 0$

$\sf→ t(t + 24) -4( t + 24)= 0$

$\sf→ (t - 4)( t + 24)= 0$

$\sf→t = 4 \\ \: \: \: \: \: \: \sf→t = - 24$

We know that Time Can't be Negative So

• Time = 4 second