Question

A force of 1500 N is acting on a vehicle horizontally of mass 200 kg and the vehicle starts fron rest.What will be the speed of vehicle when iy covers a distance of 30 m?
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Answer 1

Answer:

22.5km

Explanation:

1500N÷200kg=7.5

Now,

30×7.5=22.5km/h

Answer 2

As per the data given in the above question

we have to find the speed of the vehicle .

Given

Force ,F=1500 N

mass,m= 200 kg

distance,s= 30 m

initial velocity =0

Now ,

$Force = mass \times acceleration </p><p></p><p></p><p></p><p>$

$F=ma$

$1500=200 \times a$

Shift the 200 the get the value of a,

$\frac{1500}{200} = a$

$\frac{15}{2} = a$

$a = 7.5 \: \frac{m}{ {s}^{2} }$

• when the vehicle start from rest it means

initial speed ,u=0

Thus, from the Second equation of motion.

$s= u t+ \frac{1}{2} a {t}^{2}$

s=30 m, u=0 and a= 7.5

From this we can find the time ,

$30= 0 \times t+ \frac{1}{2} \times 7.5 \times {t}^{2}$

$30= 0+ \frac{1}{2} \times 7.5 \times {t}^{2}$

$30= \frac{7.5}{2} \times {t}^{2}$

shift the values to get t individual,

${t}^{2} = \frac{30 \times 2}{7.5}$

${t}^{2} = \frac{60}{7.5}$

${t}^{2} = 8sec$

$t = \sqrt{8} = 2.8 \: sec$

Then we have to find the speed ,

$speed = \frac{distance}{time}$

$speed = \frac{30}{2.8}$

$speed = 10.7 \: \frac{m}{s}$

Hence ,speed is 10.7 ms-¹.

Project code #SPJ2