Question

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 28 N acts on a certain object, the

acceleration of the object is 4 m/s. If the force is changed to 49 N, what will be the acceleration of the object?

Answer 1

`\\ \tt\longmapsto Force\propto Acceleration`

• F_1=28N
• a_1=4m/s^2
• F_2=49N
• a_2=?

`\\ \tt\longmapsto F_1a_2=F_2a_1`

`\\ \tt\longmapsto 28a_2=4(49)`

`\\ \tt\longmapsto 28a_2=196`

`\\ \tt\longmapsto a_2=7m/s^2`

Answer 2

• 7 m/s

Explanation:

Direct variation equation:

• y = kx, where y- force, x- acceleration, k - coefficient

We have:

• y = 28 and x = 4

Find the value of k:

• 28 = 4k
• k = 28/4
• k = 7

Our equation is:

• y = 7x

Now, find the value of x when y = 49:

• 49 = 7x
• x = 49/7
• x = 7