Question

3. A car has initial velocity of 72 km/h. It is accelerated
at 2 m/s2. Calculate the final velocity and the
distance covered after 3 seconds.​

Answer 1

Answer:

We know that a= (v-u)/t. Where a is the acceleration of an object, v is the final velocity of an object, u is the initial velocity of an object and t is the time taken to cover the distance by the object. Now given a= 2m/s2 , u= 72km/ hr, v= ?,

t= 3seconds. Here we should convert u= 72km/hr into metres/ second i.e u = 72× 5/18= 20m/s. Now, a= (v-u)/t, 2= (v-20)/3 i.e 2×3= v-20, 6 = v-20 , v= 20+6, v= 26m/s, The distance covered after 3 seconds is ? We know that s=d/ t, s=26m/s Distance covered after 3 seconds Hence we took final velocity = v = 26m/s. t= 3seconds, Now d= St, d=26×3 , d= 78m. Hence the final velocity of the car after 3 seconds is 26m/s and distance covered by the car after 3 seconds is 78m. Hope this helps you.

Answer 2

Answer:

It the question of 9th standard

Explanation:

as we know

acceleration (a) = final velocity (v) -initial velocity(u)/time.

final velocity=26m/s

distance = 87m

…hope this helps you…