Physics acceleration questions please help
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Physics acceleration questions please help

[From: ] [author: ] [Date: 11-08-15] [Hit: ]
so that the two cars meet for the first time at the next exit, which is 1.0 km away?2.)Starting from rest, a speedboat reaches a speed of 4.......
1.)A car is traveling at a constant speed of 16 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 1.0 km away?

2.)Starting from rest, a speedboat reaches a speed of 4.2 m/s in 2.5 s. What is the boat's speed after an additional 3.6 s has elapsed, assuming the boat's acceleration remains the same?

i know its a lot but i have to do them for a class online, and i haven't learned any acceleration yet and knowing how to do these would help with all the others.

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1)
t = D/v1 = 1000/16 = 62.5 sec
D = ½a*t² → a = 2D/t² = 2*1000/62.5²

a = .512 m/s²

2)
a = dv/dt = 4.2/2.5 = 1.68 m/s²
V = a*t;
V3.6 = 1.68*(2.5+3.6) = 10.248 m/s

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1.) the driving car; x(t) = v * t, so it will take t = x/4 to get to the next ramp (i will fill in the numbers later)
the accelarting car x = 1/2 * a * t^2 ---> a = 2x/t^2 is the acceleration needed,
since t = x/v, a = 2x/(x/v)^2 = 2x * v^2 / x^2 = 2(v^2)/x. with v = 16m/s, and x = 1000m, a = 2*(16^2)/1000 = 0.512 m/s^2
also nothe that the dimensions of v^2/x are [meter] [second]^-2 as they are supposed to

2.)
v = a * t, so a = v/t. given the 2.5s and the 4.2 m/s, a = 4.2/2.5 = 1.68 m/s^-2
2.5+3.6 = 6.1 seconds total, so the final speed v_f = a * t_t = 1.68 * 6.1 = 10.248 m/s
1
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