Question

On observing a sign board about speed limit, a driver applies brakes and reduces his speed from 100 m/s to 50 m/s though a distance of 50 m. The uniform retardation provided by the brakes (in m/s2) will be

♡ 25

♡ 75

♡ 50

♡ 10​

Answer 1

Answer:

$\bigstar{\bold{Option\:B:-75\:m/s^{2} }}$

Explanation:

$\Large{\underline{\rm{Given:}}}$

• Initial velocity = 100 m/s
• Final velocity = 50 m/s
• Distance travelled = 50 m

$\Large{\underline{\rm{To\:Find:}}}$

• Retardation

$\Large{\underline{\rm{Solution:}}}$

➜ Here we have to find the deceleration of the vehicle.

➜ By Newton's third equation of motion we know that,

    v² - u² = 2as

    where v = final velocity

               u = initial velocity

               a = acceleration

               s = distance travelled

➜ Substitute the data,

     50² - 100² = 2 × a × 50

     100 a = 2500 - 10000

     100 a = -7500

     a = -7500/100

    a = -75

➜ Hence the retardation of the vehicle is -75 m/s²

➜ Here acceleration is negative since it is retardation or deceleration.

➜ Hence option B is correct.

    $\boxed{\bold{Retardation=-75\:m/s^{2} }}$

$\Large{\underline{\rm{Notes:}}}$

➜ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as

Answer 2

Answer:

a= -75

Explanation:

'-' indicates retardation.