Body moving at 20 metres per second undergoes acceleration of 4 metre per second square what is speed and distance covered after 4 seconds
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Answered by
2
it is not mentioned in the question that whether the acceleration is uniform.
So i assume it to be uniform and the body is moving in a staright line.
v(final )=v(initial) + acceleration ×time
v(final)= 20+4×4=36
now, speed =magnitude of(velocity)
=36
distance=speed ×time
=36×4=144
So i assume it to be uniform and the body is moving in a staright line.
v(final )=v(initial) + acceleration ×time
v(final)= 20+4×4=36
now, speed =magnitude of(velocity)
=36
distance=speed ×time
=36×4=144
Answered by
0
u= 20m/s
a= 4 m/s²
t= 4 sec
a = v- u/t
4 = v-20/4
16 = v-20
v= 16+20= 36 m/s
s = ut +1/2 at²
= 20×4 + 1/2 * 4 × 4²
= 80+ 2×16
s= 80+32
s= 112m
a= 4 m/s²
t= 4 sec
a = v- u/t
4 = v-20/4
16 = v-20
v= 16+20= 36 m/s
s = ut +1/2 at²
= 20×4 + 1/2 * 4 × 4²
= 80+ 2×16
s= 80+32
s= 112m
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