Question

A body starting with an initial velocity of 4m/s move with a uniform acceleration of 0.5m/s². After sometime its velocity becomes 6m/s. Calculate the time taken by the body to attend this velocity and the distance covered by it during this time interval?​

Answer 1

Answer:

Provided that:

• Initial velocity = 4 m/s
• Acceleration = 0.5 m/s²
• Final velocity = 6 m/s

To calculate:

• Time taken
• Distance travelled

Solution:

• Time taken = 4 s
• Distance travelled = 18 m

Using concepts:

• First equation of motion
• Second equation of motion

Using formulas:

First equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$

Second equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}$

Where, a denotes acceleration, u denotes initial velocity, v denotes final velocity, s denotes displacement or distance or height and t denotes time taken.

Required solution:

~ Firstly let us calculate the time taken by using first equation of motion!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 6 = 4 + 0.5(t) \\ \\ :\implies \sf 6 - 4 = 0.5t \\ \\ :\implies \sf 2 = 0.5t \\ \\ :\implies \sf \dfrac{2}{0.5} \: = t \\ \\ :\implies \sf \dfrac{20}{5} \: = t \\ \\ :\implies \sf 4 \: = t \\ \\ :\implies \sf t \: = 4 \: seconds \\ \\ :\implies \sf Time \: = 4 \: seconds$

• Henceforth, time = 4 sec!

~ Now let's calculate distance by using second equation of motion!

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 4(5) + \dfrac{1}{2} \times 0.5(4)^{2} \\ \\ :\implies \sf s \: = 4(5) + \dfrac{1}{2} \times 0.5 \times 16 \\ \\ :\implies \sf s \: = 20 + \dfrac{1}{2} \times 0.5 \times 16 \\ \\ :\implies \sf s \: = 10 + 1 \times 0.5 \times 16 \\ \\ :\implies \sf s \: = 10 + 1 \times 8 \\ \\ :\implies \sf s \: = 10 + 8 \\ \\ :\implies \sf s \: = 18 \: m \\ \\ :\implies \sf Distance \: = 18 \: m$

• Henceforth, distance = 18 m