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)

Answers

Answer 1
The velocity of the car after 10 s is 78.95 km/hr

Explanation:

Given:

m = 1,250 kg

[tex]v_i[/tex] = 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 = [tex]\frac{v_f \:-\:v_i}{t}[/tex]

where: a - acceleration

[tex]v_i[/tex] - initial velocity

[tex]v_f[/tex] - 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 = [tex]\frac{1700\:N}{1250\:N}[/tex]

a = 1.36 m/s²

Solve for final velocity using the formula for accelerationConvert 30 km/hr to m/s

= [tex]\frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}[/tex]

= [tex] 8.33 m/s [/tex]

Substitute the value of a, [tex]v_i[/tex] and t

a = [tex]\frac{v_f \:-\:v_i}{t}[/tex]

[tex]1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}[/tex]

[tex](10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s[/tex]

[tex]v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s[/tex]

[tex]v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s[/tex]

[tex]v_f\: =\: 21.93\: m/s[/tex]

Convert to km/hr

= [tex]\frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}[/tex]

= [tex]78.95\: km/hr[/tex]

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

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