Question

A body starting from rest covers a distance of 40 m in 4s with constant
acceleration along a straight line. Compute its final velocity and the time
required to cover half of the total distance.

Answer 1

Answer:

• u = 0
• s = 40m
• T = 4 s
• s = ut + 1/2 at2
• 40= 0 + 1/2 ×16×a
• 40 = 0 + 8a
• 40/8 = a
• 5 = a

v= u+ at

v= 0+ 5× 4

v = 20

Answer 2

Explanation:

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ u = 0 \\ s = 40m \\ t = 4s \\ \\ \frac{40 \times 2}{4 \times 4} = a = 5 {ms}^{ - 2}$

And Now according to the Question For Half Distance

$as \: mention \: acceleration \: is \: constant \: \\ means \: not \: change = 5 {ms}^{ - 2} \\ \\ again \\ s = \frac{1}{2} a {t}^{2} \\ s = 20m(half) \\ \\ \ \sqrt{ \frac{20 \times 2}{5} } = t = 2 \sqrt{2} \: s \\ \\ v = u + at \\ \\ v = 5 \times 2 \sqrt{2} = 10 \sqrt{2} \: {ms}^{ - 1}$

Hope it will help.

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