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 ?

Explanation

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):})`
`a 4 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`