A particle move rectilinearly with a constant acceleration of 1m/s

^{2}.its speed after 10second is 5 m/s .The distance cover by particle in this duration is ? ( intial and final velocities are in opposite direction )
Tutors, please sign in to answer this question.

Medfield, MA

If it started from 0 and accelerated at 1 m/s/s for 10 s how fast would it be going?

=> 10 m/s

So it must have started from a negative speed of -5 m/s to get to 5 m/s after 10 sec.

Are you sure it is asking for distance and not displacement?

Start at a point, go for 5 sec negative while slowing to 0 m/s,

then 5 sec positive speeding up to 5 m/s

And you get back to where you started => displacement is zero.

The distance traveled is from the formula x_{f} = x_{0} + v_{0}t + 1/2(at^{2})

but you need to do it in 2 parts

SO

x = 1/2(at^{2}) = 1/2(1 m/s/s) (5 s)^{2}

= 12.5 m in negative direction

them another 12.5 meters coming back

= 25 meters distance traveled.

- Math 9857
- Math Help 5445
- Word Problem 5117
- Algebra 5057
- Math Word Problem 4500
- Algebra 1 4075
- Algebra 2 3503
- Algebra Word Problem 2482
- Chemistry 2250
- Calculus 2205