Question

A bus starts from rest accelerates in a straight line at a constant rate 3 ms-2 for 8 seconds. Calculate the distance travelled by the bus during this time interval.

Answer 1

your ans

bus starts from rest

so

u=0

a=3

t=8

to find distance

s=ut +1/2at2

s=0×8+1/2×3×8×8

s=96m

Answer 2

$\bf \large \it{Hey \: User!!!}$

according to the question, the bus starts from rest and accelerates in a straight line at a constant rate 3ms^-2 for 8 seconds.

the bus starts from rest, therefore it's initial velocity (u) = 0ms^-1

acceleration = 3ms^-2

time = 8 seconds

we can find it's distance by both two second and third equation.

by second equation :-

>> s = ut + 1/2 at²
>> s = (0×8) + 1/2 (3) (8)²
>> s = 0 + 3/2 × 64
>> s = 3 × 32
>> s = 96m

for 3rd equation, we need to find it's final velocity.

by first equation of motion, we get :-

>> v = u + at
>> v = 0 + ( 3 × 8 )
>> v = 24ms^-1

the final velocity of the bus is 24ms^-1

now, we have to find the distance (s) travelled by the bus in 8 seconds.

>> 2as = v² - u²
>> 2 (3) (s) = 24² - 0²
>> 6s = 576
>> s = 576/6
>> s = 96m

hence, the distance travelled by the bus in 8 seconds is 96m.

$\bf \large \it{Cheers!!!}$