Question

A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after 10 seconds? (Neglect friction)

Answer 1

The velocity of the car after 10 s is 78.95 km/hr

Step-by-step explanation:

Given:

m = 1,250 kg

`v_i` = 30 km/hr

F = 1,700 N

t = 10 s

Required:

Final velocity

Equation:

Force

F = ma

where: F - force

m - mass

a - acceleration

Acceleration

a =
`(v_f \:-\:v_i)/(t)`

where: a - acceleration

`v_i` - initial velocity

`v_f` - final velocity

t - time elapsed

Solution:

Solve for acceleration using the formula for force

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a =
`(1700\:N)/(1250\:N)`

a = 1.36 m/s²

Solve for final velocity using the formula for acceleration

• Convert 30 km/hr to m/s

=
`(30\:km)/(hr)\:×\:(1000\:m)/(1\:m)\:×\:(1\:hr)/(3600\:s)`

=
`8.33 m/s`

• Substitute the value of a,
  `v_i` and t

a =
`(v_f \:-\:v_i)/(t)`

`1.36\: m/s² \:= \:(v_f \:-\:8.33\:m/s)/(10\:s)`

`(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s`

`v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s`

`v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s`

`v_f\: =\: 21.93\: m/s`

• Convert to km/hr

=
`(21.93\:m)/(s)\:×\:(1\:km)/(1000\:m)\:×\:(3600\:s)/(1/:hr)`

=
`78.95\: km/hr`

Final answer

The velocity of the car after 10 s is 78.95 km/hr