A particle located initially at -7.00 m accelerates from rest to a velocity of 2

Question

A particle located initially at -7.00 m accelerates from rest to a velocity of 2.50 m/s in a time interval of 1.10 s.
Calculate the acceleration of the particle. (3 SF)

HINTS: What is the initial velocity if the particle starts out "at rest"? Don't make this part harder than it needs to be. Use only the information you need to find the acceleration.

Calculate the distance the particle moves during the 1.10 s time interval. (3 SF)

HINTS: Notice that this problem asks for the distance travelled, which is really the displacement, x. If you use xi in your calculation, then xf will be the final position, and not the distance traveled.

Explanation / Answer

A) Initial Velocity U = 0 (because starts from rest), final velocity V = 2.5 m/sec. Change in Velocity dV = V-U = 2.5 Time taken dt = 1.1 seconds. So acceleration a =dv/dt = 2.5/1.1 = 2.273 m/sec2.

B) Distance travelled S = ut + 1/2.a.t2 = 0 + [0.5x2.273x(1.1)2] = 1.375 m.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.