A) 10 m
B) 20 m
C) 40 m
D) 5m
Correct Answer: B
Solution :
Rate of change of speed gives acceleration. \[\therefore \]Acceleration/Retardation \[=\frac{change\text{ }in\text{ }speed}{time}\] \[\therefore \]Retardation\[(a)=\frac{30-10}{5}=-4\,m/{{s}^{2}}\] From equation of motion: when \[t=2s,\] \[{{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}\] \[=(30\times 2)-\frac{1}{2}\times 4\times {{(2)}^{2}}=52\,m\] When \[t=3\text{ }s,\] \[{{s}_{2}}=30\times 3-\frac{1}{2}\times 4{{(3)}^{2}}=72\,cm\] \[\therefore \] \[{{s}_{2}}-{{s}_{1}}=72-52=20\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
