What acceleration must the car maintain so they meet at the next exit?
slowphysicsstudent asked:
A car is traveling at a constant speed of 34 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 1.4 km away?
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A car is traveling at a constant speed of 34 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 1.4 km away?
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April 22nd, 2009 at 8:36 am
The first car would take (1400/34 seconds) = 41.18 sec to cover the 1.4 kms with constant velocity of 34m/s.
Now the other car entering from ramp should also travel the same distance in the same time to meet the previous car.
Let the acceleration be a m/s^2.
Then s= ut + 1/2 *a *t^2…………here intial velocity u = 0
s = 1/2 *a *t^2
a = 2 *s/t^2
= 2 * 1400 / (41.18 )^2
=1.65 m/s^2 is the required answer.