deceleration braking-force

Question

3.3 A motorbike is travelling on a horizontal road at a velocity of 60 km/h. The mass of the motorbike is 800 kg and the resistance to motion is 510 N. The motorbike stops over a distance of 40 m when the brakes are applied. Calculate the following: 3.3.1 The deceleration of the motorbike 3.3.2 The braking force of the motorbike

Answer 1

3.3.1)

Convert speed of the motorbike from km/h to m/s.

u =( 60)(5/18)  m/s

   = 16.667 m/s

Find the deceleration of the of the motor bike by using the basic dynamic equation.

v² - u² = 2as

Here, u is the initial velocity, v is the final velocity, and s is the distance.

Substitute 0 for v , 16.667 m/s for v , and 40 m for s

0 - 16.667² =2a(40)

a = - 16.667²/ 80

   = 6.94 m/s²

So deceleration of the motor bike is 6.94 m/s².

Answer 2

3.3.2)

Convert speed of the motorbike from km/h to m/s.

u =( 60)(5/18)  m/s

   = 16.667 m/s

Find the deceleration of the of the motor bike by using the basic dynamic equation.

v² - u² = 2as

Here, u is the initial velocity, v is the final velocity, and s is the distance.

Substitute 0 for v , 16.667 m/s for v , and 40 m for s

0 - 16.667² =2a(40)

a = - 16.667²/ 80

   = - 6.94 m/s²

So deceleration of the motor bike is 6.94 m/s².

Find the breaking force by considering equilibrium

∑F = ma

F_R + F_B  = - ma

510  +  F_B  = - (800)(-6.94)

510  +  F_B  = 5555.55

F_B  = 5045.55 N