| The velocity-time graphs of a car and a scooter are shown in the figure. (i) the difference between the distance travelled by the car and the scooter in 15s and (ii) the time at which the car will catch up with the scooter are, respectively [JEE Online 15-04-2018] |
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A) \[337.5m\,\,\,and\,\,25s\]
B) \[225.5m\,\,and\,\,10s\]
C) \[112.5m\,\,and\,\,22.5s\]
D) \[11.2.5m\,\,and\,\,15s\]
Correct Answer: C
Solution :
| [c] The distance travelled in 15 seconds by both will be given by area under curve. |
| For car, \[{{s}_{1}}=\frac{1}{2}\times 15\times 45\] |
| \[{{s}_{1}}=337.5m\] |
| For scooter, \[{{s}_{2}}=15\times 30\] |
| \[{{s}_{2}}=450\] |
| \[{{s}_{1}}-{{s}_{2}}=112.5m\] |
| Suppose after time t, they will meet, then distance travelled by both of them will be equal. |
| \[30t=\frac{1}{2}\times 15\times 45+45\times (t-15)\] |
| \[30t=337.5+45t-675\] |
| \[15t=337.5\] |
| \[t=22.5s\] |
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