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

Question

2 years ago

13

10 seconds? (Neglect friction)

1 answer:

Talja [164] · Answer 1 · 2022-06-23T11:13:02+03:00

<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

$v_i$ = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = $\frac{v_f \:-\:v_i}{t}$

where: a - acceleration

$v_i$ - initial velocity

$v_f$ - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

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

a = $\frac{1700\:N}{1250\:N}$

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>

= $\frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}$

= $8.33 m/s$

a = $\frac{v_f \:-\:v_i}{t}$

$1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}$

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

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

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

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

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

= $78.95\: km/hr$

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>