mechanic q

found acceleration, (2t-3)i -12j

put int in F=MA

6.5=0.5 * (2t-3)i -12j

i tried to find magnitude which is wong, was next

Reply 1

aaasd

Ur finding magnitude so

(6.5/0.5)^2 = (2t-3)^2 + (-12)^2

Then solve the quadratic to get t=4 or -1

Reply 2

jeriu

but why divide by 0.5, can u do it my way, cah it looks like u just copied MS

Original post by aaasd

Ur finding magnitude so

(6.5/0.5)^2 = (2t-3)^2 + (-12)^2

Then solve the quadratic to get t=4 or -1

Reply 3

hggkhghk

f=ma
6.5=0.5(2ti-3j)

but why divide by 0.5, can u do it my way, cah it looks like u just copied MS

Reply 4

mqb2766

If is wrong to write this as you have a scalar on the left and a vector on the right.

It is nearly right (6.5=0.5*(2t-3)i-12j), but you're equating magnitudes as you're given the magnitude of the force, so you have to take the magnitude of the right hand side, as done in Post 2.

Original post by hggkhghk

f=ma
6.5=0.5(2ti-3j)

(edited 4 years ago)

Reply 5

jeriu

wait , sorry, so isn't the LHS a force which is a magnitude ?!?

Original post by mqb2766

Reply 6

mqb2766

The given force is the magnitude (scalar).
You need to find the magnitude of the vector acceleration and then equate (using the mass).

Reply 7

jeriu

also surely need to sqrt

Original post by mqb2766

The given force is the magnitude (scalar).
You need to find the magnitude of the vector acceleration and then equate (using the mass).

Reply 8

mqb2766

When a is the scalar acceleration
6.5 = ma
6.5^2 / m^2 = a^2
Post 2 had the correct solution.

Reply 9

mysticmagic2000

Slightly different method, but as f=má you can work out that the acceleration would be 13m/s^2
Differentiate the v=(t^2-3t)i -12tj to get a= (2t-3)i -12j. As you know 13 is the magnitude, write square root of (2t-3)^2 + (12)^2 is equal to 13. Therefore. (2t-3)^2= 25. Square root that 2t-3=5. 2t=8 t=4

Original post by jeriu