Question

a docs.google.com
Section 5
A body moving with an initial velocity 10 m/s is brought to rest over a
distance of 50 m. The time taken for the same is:
O 4s
O 6s
O8s
О10s

​

Answer 1

10 s is the required answer.

GIVEN:

• Initial velocity of the body, u = 10 m/s
• Distance, s = 50 m
• Since the body is brought to rest, its final velocity is zero. v = 0

TO FIND:

Time taken for the body to stop, t

FORMULAE:

• According to third equation of motion, v² - u² = 2as

• According to first equation of motion, v = u + at

SOLUTION:

STEP 1: TO FIND ACCELERATION

${v}^{2} - {u}^{2} = 2as \\ \\ {0}^{2} - {10}^{2} = 2 \times a \times 50 \\ \\ - 100 = 100 \times a \\ \\ a = - \frac{100}{100} \\ \\ a = - 1 \: \frac{m}{ {s}^{2} }$

The acceleration is -1 m/s².

Negative sign indicates the retardation of the body.

STEP 2: TO FIND TIME

$v = u + at \\ \\ 0 = 10 + ( - 1\times \: t ) \\ \\ 0 = 10 - t \\ \\ t = 10 \: seconds$

ANSWER:

Time taken for the body to stop is 10 seconds.

OTHER FORMULA:

The time taken can also be found by using second equation of motion, instead of first equation of motion.

According to second equation of motion, s = ut + (at²)/2

$s = ut + \dfrac{1}{2} a {t}^{2} \\ \\ 50 =( 10 \times t) + ( \dfrac{1}{2} \times ( - 1) \times {t}^{2} ) \\ \\ 50 = 10t - \dfrac{ {t}^{2} }{2} \\ \\ 50 = \dfrac{(2 \times 10t )- {t}^{2} }{2} \\ \\ 50 \times 2 = 20t - {t}^{2} \\ \\ 100 = 20t - {t}^{2} \\ \\ {t}^{2} - 20t + 100 = 0 \\ \\ {t}^{2} - 10t - 10t + 100 = 0 \\ \\ t(t - 10) - 10(t - 10) = 0 \\ \\ (t - 10)(t - 10) = 0 \\ \\ t - 10 = 0 \\ \\ t = 10 \: s$