Question

A car stops 4 seconds after the application of the brakes while covering a rectilinear stretch 337 feet long. If the motion occurred with a constant acceleration a_{c} determine the initial speed..??​

Answer 1

Answer:

337 ft/s

Explanation:

Final velocity = v = 0 ( as the body is at rest finally)

displacement covered = s = 337ft

acceleration = a = constant

initial velocity = u

time taken = t = 4s

we know,

$s = ut + (a {t}^{2} \div 2)$

thus, here, 337 = 4u + 8a (I)

also,

${v}^{2} - {u}^{2} = 2as$

So, - u² = 674a

a = - u²/674

Putting value of a in (I),

we get

$( - 4{u}^{2} \div 337) + 4u - 337 = 0$

so,

$- 4 {u}^{2} + 1348 - 1 = 0$

so,

$4 {u}^{2} - 1348 + 1 = 0$

Using quadratic formula,

$u = 337ft {s}^{ - 1}$

also,

$a = 168.5ft {s}^{ - 2}$

Answer 2

Answer:

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