Question

a bus was moving with a speed of 54km/h on applying brakes it stopped in 8 seconds calculate the acceleration and the distance traveled before stopping​

Answer 1

GIVEN:

• Initial velocity of the bus , u = 54 km/hr

• Time at which it stops, t = 8 sec

• Since the bus stops after applying brake, it's finaly velocity is zero. v = 0

TO FIND:

• Acceleration of bus, a

• Distance travelled before stopping , s

FORMULAE:

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

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

SOLUTION:

STEP 1: TO CONVERT INITIAL VELOCITY IN STANDARD UNIT:

• SI unit of velocity is m/s

• 1 km/hr = 5/18 m/s

• 54 km/hr = 54 × (5/18) m/s = 3 × 5 m/s

• Initial velocity , v = 15 m/s

STEP 2: TO FIND ACCELERATION:

v = u + at

0 = 15 + 8a

-15 = 8a

a = -15/8

a = -1.875 m/s²

Here negative sign indicates the retardation.

STEP 3: TO FIND DISTANCE:

${v}^{2} - {u}^{2} = 2as \\ \\ 0 - {15}^{2} = 2( \frac{ - 15}{8} )s \\ \\ - 15 \times 15 = - (\frac{15}{4} )s \\ \\ s = \frac{15 \times 15 \times 4}{15} \\ \\ s = 15 \times 4 \\ \\ s = 60 \: m$

ANSWER:

• Acceleration of bus is -1.875 m/s²

• Distance travelled before stopping is 60 m.