Question

a motorcycle moving with a speed of 5 metre per second is subjected to an acceleration of 0.2 m/sec square calculate the speed of motorcycle after 10 second and the distance travelled in the this answer​

Answer 1

Initial Velocity = 5 m/sec

Acceleration = 0.2 m/sec²

Time given =10 seconds

To Calculate :-

(i) Final velocity after 10 seconds.

(ii) Distance covered in 10 seconds.

$\rule{200}{2}$

From first equation of motion we have :-

$\boxed{ \red{ \bold{ \sf{ \large{v = u + at}}}}}$

Thus,

Final velocity after 10 seconds :-

$v = 5 + 0.2 \times 10 \\ \\ \\ v =5 + 2 = 7$

Final velocity = 7 m/sec

$\rule{200}{2}$

Also we know second equation of motion:-

$\boxed{ \bold{ \red{ \large{s = ut + \frac{1}{2}a {t}^{2}}}}}$

Applying this :-

$s = 5 \times 10 + \frac{1}{2} \times 0.2 \times 100 \\ \\ \\ = 50 + 10 = 60m$

Distance travelled in 10 seconds is equal to 60 metres.

$\rule{200}{2}$

Answer 2

Answer:

initial velocity =5m/s

acceleration =0.2m/s2

time=10s

v-u=at

v-5=0.2×10

v=2+5=7m/s

s=ut+1/2at2

s=5×10+1/2×0.2×10×10

s=50+10=60m