Question

a running car has a uniform acceleration of 4ms^-2. what distance it will cover in 10 secs after the start?​

Answer 1

Answer:

s = ut + 1/2 a$t^{2}$

s =  1/2 x 4 x 10x10            (u= 0 )

distance = 200m

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Answer 2

Answer:

Given:

Given: Initial velocity is: u=0

Given: Initial velocity is: u=0Acceleration is: a=4 ms

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 s

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 21

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 21

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 21 ×4×10

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 21 ×4×10 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 21 ×4×10 2

Given: Initial velocity is: u=0Acceleration is: a=4 ms 2 Time is: t=10 sUsing second equation of motion:s=ut+ 21 at 2 s=0×10+ 21 ×4×10 2 =0+200=200 m

Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration.