Question

A body A starts from rest and accelerates at constant rate x for time t. Another body B moves with constant speed x for the time t. Compare the displacement if A and B


pls show me the whole process ​

Answer 1

Answer:

I don't know your answer

Explanation:

I am so sorry

Answer 2

The displacement of A is t/2 times the displacement of B

Explanation:

For body A

Initial velocity $u_A=0$

Acceleration $a_A=x$

In time t the displacement of the body A

By the second equation of motion

$s_A=u_At+\frac{1}{2}a_At^2$

$\implies s_A=0+\frac{1}{2}\times x\times t^2$

$\implies s_A=\frac{xt^2}{2}$

Body B  moves with constant speed x for the time t.

Therefore, the displacement of body B in time t

$s_B=x\times t$

$\implies s_B=xt$

$\frac{s_A}{s_B}=\frac{xt^2/2}{xt}$

$\implies \frac{s_A}{s_B}=\frac{t}{2}$

$\implies s_A=s_B\times \frac{t}{2}$

Therefore, the displacement of A is t/2 times the displacement of B

Hope this answer is helpful.

Know More:

Q: A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed 18 km/h find (a) the average velocity during this period , and (b) the distance traveled by the particle during this period

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