A car accelerated from rest at 2m/s^2. What will be its speed and how far will be from its starting point after 10s? After what distance will its speed be 40m/s? (Please send me the solution with an attachment)
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Assuming that
a(t)=dv/dt
Then you can do the substitution,
a(t)=dv/ds∗ds/dt=(v∗dv)/ds
From this, multiply with ds and integrate as we can say in physics (formally we are doing an integration on both sides, but that would give the same result). This gives us
∫a∗ds=∫v∗dv
since a is a function of t it is not affected by the integration and can be moved out from the integration,
a∫ds=∫v∗dv
and this gives,
as=v2/2−v02/2
we know that v=0 and v0=20. We also know s=10. Using these values in the final equation we get,
a∗10=02/2−202/2
a∗10=−200<=>a=−20m/s
since we only have 2 measurement points, this will be the average acceleration. The true acceleration might not be linear, but I would assume quite close to.
a(t)=dv/dt
Then you can do the substitution,
a(t)=dv/ds∗ds/dt=(v∗dv)/ds
From this, multiply with ds and integrate as we can say in physics (formally we are doing an integration on both sides, but that would give the same result). This gives us
∫a∗ds=∫v∗dv
since a is a function of t it is not affected by the integration and can be moved out from the integration,
a∫ds=∫v∗dv
and this gives,
as=v2/2−v02/2
we know that v=0 and v0=20. We also know s=10. Using these values in the final equation we get,
a∗10=02/2−202/2
a∗10=−200<=>a=−20m/s
since we only have 2 measurement points, this will be the average acceleration. The true acceleration might not be linear, but I would assume quite close to.
Answered by
2
Using Newton's equation for uniformly accelerated motion:
S=ut+1/2at²
u=0
t=2 sec.
a=2m/s²
S= 0×2+1/2×2×2²=0+4=4
Hence distance covered by the body in 2 seconds is just 4 meters.
Ans. 4m.
S=ut+1/2at²
u=0
t=2 sec.
a=2m/s²
S= 0×2+1/2×2×2²=0+4=4
Hence distance covered by the body in 2 seconds is just 4 meters.
Ans. 4m.
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