An object is moving with initial velocity of 20m/s. It must slow down to a stop in .33 seconds and travel 4m. Write a solution that applies a force to the object to satisfy the expected result.?

Question

The first thing I do not understand is why I can assume linear acceleration.
I have a basic understanding of calculus so explaining using integration is fine.

Elena Albarran · Accepted Answer

#F = m * 36.7 m/s^2#

You may “assume” linear acceleration because you are asked for “a solution”, not all solutions.

F = m * a d = v * t To stop in the specified distance and time. 4m = v(m/s) * 0.33s ; v = 12.12 m/s

To decelerate: #a = (v/t) ; a = ((12.12(m/s))/0.33s) = 36.7 m/s^2#

F = m * a

#F = m * 36.7 m/s^2#

Isabella Alvarez · Answer

Applied force #F(t)=(-22.0m-233.7mt)# N

Initial velocity #u=20ms^-1#. Final velocity #v=0#, distance traveled #=4m#. The applicable kinematic equation for a constant acceleration is #v^2-u^2=2as# #=>0^2-20^2=2axx4# #=>a=-400/8=-50ms^-2# Let us find out for what time this retarding force is applied. Using the other kinematic equation #v=u+at# we get #0=20+(-50)t# #t=20/50=0.4s# We see that the object takes more time than given in the question to come to stop. This implies that it is not a case of constant deceleration. Let us assume that the object of mass #m# is acted upon by a time dependent force #F(t)#. From Newton's Second law we get #F(t)=mxxa(t)# .....(1) We know that acceleration can be written as #(dv)/dt=a(t)# #=>dv=a(t)cdot dt# Using (1) and Integrating both sides we get #v(t)=int(F(t))/mcdot dt#......(2) Suppose the force is given by the expression #F(t)=L+Kt# Where #L and K# are constants. (2) becomes #v(t)=int (L+Kt)/mcdot dt# #=>v(t)= L/m t+K/mt^2/2+C# where #C# is constant of integration. As #v(t)=u# initial velocity at #t=0# Expression becomes #v(t)=L/mt+K/(2m)t^2+u# .....(3) We also know that velocity can be written in terms of displacement #s# as #(ds)/dt=v(t)# #=>ds=v(t)cdot dt# Integrating both sides we get #s=intv(t)cdot dt# Using (3) we get #s=int(L/mt+K/(2m)t^2+u)cdot dt# #=>s=L/mt^2/2+K/(2m)t^3/3+ut+C_1# #C_1# is constant of integration. Assuming that displacement is zero at #t=0#, we get #C_1=0# and the expression for displacement becomes #=>s=L/(2m)t^2+K/(6m)t^3+ut# ......(4) Inserting given values in (3) and (4) we obtain #0=L/mxx0.33+K/(2m)(0.33)^2+20# .....(5) #4=L/(2m)(0.33)^2+K/(6m)(0.33)^3+20xx0.33# ......(6) To solve for #L and K#, multiply (5) with #0.33/2#, it becomes #0=L/(2m)xx(0.33)^2+K/(4m)(0.33)^3+20xx0.33/2# .....(7) Subtracting (7) from (6) #4=K/(6m)(0.33)^3+20xx0.33-(K/(4m)(0.33)^3+20xx0.33/2)# #=>4=-K/(12m)(0.33)^3-20xx0.33/2# #K=-233.7m# Inserting this value of #K# in (5) we get #0=L/mxx0.33-(233.7m)/(2m)(0.33)^2+20# #L=-22.0m# As such applied force #F(t)=(-22.0m-233.7mt)# N

Benjamin Asiedu · Answer

undefined To bring the object to a stop in 0.33 seconds while traveling 4m, a force needs to be applied to decelerate it. Using the equation $v_f = v_i + at$, where $v_f$ is final velocity (0 m/s), $v_i$ is initial velocity (20 m/s), $a$ is acceleration, and $t$ is time, we can calculate acceleration. Subsequently, using $s = ut + \frac{1}{2}at^2$, where $s$ is displacement (4 m), $u$ is initial velocity, $a$ is acceleration, and $t$ is time, we can verify the result. The force applied would depend on the mass of the object, and $F = ma$ could be used to find it.

An object is moving with initial velocity of 20m/s. It must slow down to a stop in .33 seconds and travel 4m. Write a solution that applies a force to the object to satisfy the expected result.?

The first thing I do not understand is why I can assume linear acceleration. I have a basic understanding of calculus so explaining using integration is fine.

The first thing I do not understand is why I can assume linear acceleration.
I have a basic understanding of calculus so explaining using integration is fine.