If a car starts from rest and moves with a uniform acceleration of 10ms^-2 for ten seconds, the distance it covers in the last one second of its motion is

DrNen

1 year ago

I really need a better explanation. Anyone? 🙏I've

MAY Bby

1 year ago

Me too

Truss

1 year ago

s =1/2a(t2^2-t1^2)
i.e 1/2 × 10 (10^2-9^2)
=1/2 ×10 (100-81)
=1/2×10×19
=1/2×10×19
=190÷2
=95✓

jedroy

2 months ago

first calculate the distance covered in the full 10 seconds then calculate distance covered in 9 seconds the distance in 10 minus distance in 9 should give you the distance covered in just that last second nothing more than just what was covered in the last second if solved with the right motion eqn we should get 95m

Fortisleo3

1 year ago

the question is not well solved or explain is asking for distance per seconds