An object is moving with constant acceleration. Its velocity is 48 ms^(-1)at the end of 10 second and becomes 68 ms^(-1)at the end of 15 second. What would be the distance travelled by the object in 15 second ?

An object is moving with constant acceleration. Its velocity is 48 ms^

1.	An object is moving with constant acceleration. Its velocity is 48 ms^(-1)at the end of 10 second and becomes 68 ms^(-1)at the end of 15 second. What would be the distance travelled by the object in 15 second ?
Answer» Solution :In equation of motion `v =v _(0) + at` `t _(1) = 10 s IMPLIES v _(1) = 48 ms ^(-1)` `therefore 48 = v _(0) + 10a ""…(1) and t _(2) = 15 s implies v _(2)= 68 ms ^(-1)` `therefore 68 = v _(0) + 15 a ""…(2)` Solving equation (1) and (2), `({:(68=, v _(0)+, 15A), (48=, v_(0)+, 10 a),(-,-,-):})/({:(20=, 5A):})` `a4 ms ^(-2)` now from equation (1) `48= v _(0) + 10 a, `put `a = 4 ms ^(-2)` `therefore 48 = v _(0) + 40` `therefore v _(0) = 8 ms ^(-1)` Now displacement `d = v _(0)t + 1/2 at ^(2)` Put `v _(0) = 8 ms ^(-1), t = 15 s, a = 4 ms ^(-2)` ` therefore d = 8 xx 15 + 1/2 xx 4 xx 225` `therefore d = 120 + 450 therefore d = 570 m`