Question

A truck of mass 4000Kg was moving with a velocity of 8m/s. On application of brakes it stops after 10s. Calculate the distance travelled by it after the brakes were applied. Also find the force applied by the brakes.​

Answer 1

GIVEN:

• Mass of the truck, m = 4000 kg
• Initial velocity, u = 8 m/s
• On application of brake, the truck stops. So final velocity becomes zero. v = 0
• Time , t = 10 s

TO FIND:

• Distance travelled after the application of brake, s

• Force applied by the brakes, f

FORMULAE:

• According to Newton's second law of motion, f = ma

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

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

SOLUTION:

STEP 1: To find acceleration

$v = u + at \\ \\ 0 = 8 + (a \times 10) \\ \\ - 8 = 10a \\ \\ a = - \frac{8}{10} \\ \\ a = - 0.8 \: \frac{m}{ {s}^{2} }$

Acceleration is - 0.8 m/s².

Negative sign indicates retardation.

STEP 2: To find distance travelled.

${v}^{2} - {u}^{2} = 2as \\ \\ {0}^{2} - {8}^{2} = 2 \times ( - 0.8) \times s \\ \\ - 64 = - 1.6 \times s \\ \\ s = \frac{64}{1.6} \\ \\ s = \frac{160}{4} \\ \\ s = 40 \: m$

Distance travelled after the application of brake is 40 m.

STEP 3: To find force

$f = ma \\ \\ f = 4000 \times ( - 0.8) \\ \\ f = - 3200 \: newton$

The force applied by the brake is - 3200 N.

Negative sign indicates the retarding force.

ANSWER:

• Distance travelled after the application of brake is 40 m.

• The force applied by the brake is - 3200 N.