Question

a car of mass 500 kg increases it's velocity from 40 m/s to 60 m/s in 10 seconds. find acceleration, distance travelled and amount of force applied.​

Answer 1

Initial momentum pp1

=mu=2000kgm/s

Final momentum p 2

=mv=4000kgm/s

and time t=10s

Hence, force is F=

$= \frac{p1 + p2}{t}$

⟹F=200N

Answer 2

Answer:

Acceleration = 2 m/s²

Distance travelled = 500 m / 0.5 km

Force applied = 1000 N

Explanation:

Initial velocity of the car (u) = 40 m/s

Final velocity of the car (v) = 60 m/s

Time (t) = 10 secs

Mass of the car (m) = 500 kg

Therefore,

a = v-u/t

= 60-40/10

= 20/10 = 2 m/s²

d = ut+1/2at²

= 40×10+1/2×2×10²

= 500 m

F = ma

= 500×2

= 1000 N

hope it helps!!